Which graph represents a linear function that has a slope of 0.5 and a y-intercept of 2?
On a coordinate plane, a line goes through points (negative 2, 0) and (0, 1).
On a coordinate plane, a line goes through points (0, 2) and (4, 0).
On a coordinate plane, a line goes through points (0, 2) and (2, 3).
On a coordinate plane, a line goes through points (negative 4, 0) and (0, 2).

Answer:

Amount = Rs. 30250 when Rate = 10%

Amount = Rs. 31360 when Rate = 12%

Step-by-step explanation:

Given

[tex]Principal, P = Rs.\ 25,000[/tex]

[tex]Time, t = 2\ years[/tex]

[tex]Rate; R_1 = 10\%[/tex]

[tex]Rate; R_2 = 12\%[/tex]

Number of times (n) = Annually

[tex]n = 1[/tex]

Required

Determine the Amount for both Rates

Amount (A) is calculated by:

[tex]A = P(1 + \frac{r}{n})^{nt}[/tex]

When Rate = 10%, we have:

Substitute 25,000 for P; 2 for t; 1 for n and 10% for r

[tex]A = 25000 * (1 + \frac{10\%}{1})^{1 * 2}[/tex]

[tex]A = 25000 * (1 + \frac{10\%}{1})^{2}[/tex]

[tex]A = 25000 * (1 + 10\%)^{2}[/tex]

Convert 10% to decimal

[tex]A = 25000 * (1 + 0.10)^{2}[/tex]

[tex]A = 25000 * (1.10)^{2}[/tex]

[tex]A = 25000 * 1.21[/tex]

[tex]A = 30250[/tex]

Hence;

Amount = Rs. 30250 when Rate = 10%

When Rate = 12%, we have:

Substitute 25,000 for P; 2 for t; 1 for n and 10% for r

[tex]A = 25000 * (1 + \frac{12\%}{1})^{1 * 2}[/tex]

[tex]A = 25000 * (1 + \frac{12\%}{1})^{2}[/tex]

[tex]A = 25000 * (1 + 12\%)^{2}[/tex]

Convert 12% to decimal

[tex]A = 25000 * (1 + 0.12)^{2}[/tex]

[tex]A = 25000 * (1.12)^{2}[/tex]

[tex]A = 25000 * 1.2544[/tex]

[tex]A = 31360[/tex]

Hence;

Amount = Rs. 31360 when Rate = 12%

Answer:

p + q = -3

Step-by-step explanation:

First we need to take the original equation, and factor it to a form that's easier to get two binomial factors from (i.e., let's get a quadratic):

x^3 + px^2 + qx + 1

= x (x^2 + px + q) + 1

Now that we have factored out the x, we have a quadratic trinomial which we know can be broken down into two linear binomials.  The problem gives us two linear binomials, so let's take a look.

(x - 2) (x + 1) = (x^2 + px + q)

x^2 - 2x + x -2 = x^2 + px + q

Now let's solve.

x^2 - x - 2 = x^2 + px + q

-x - 2 = px + q

From here, we can easily see that p = -1 (the coefficient of x) and q = -2.

Hence, p + q = -1 + -2 = -3.

Cheers.

Answer:

b

Step-by-step explanation:

because its right dummy

Answer:

0

Step-by-step explanation:

Hello, dividing by 0 is not defined. so

[tex]\dfrac{2x^2}{6x}[/tex]

is defined for x different from 0

This being said, we can simplify by 2x

[tex]\dfrac{2x^2}{6x}=\dfrac{2x*x}{3*2x}=\dfrac{1}{3}x[/tex]

and this last expression is defined for any real number x.

Thank you

Step-by-step explanation:

your required answer is 60°.

Hello,

Here, in the figure;

angle 1= 120°

To find : m. of angle 2.

now,

angle 1 + angle 2= 180° { being linear pair}

or, 120° +angle 2 = 180°

or, angle 2= 180°-120°

Therefore, the measure of angle 2 is 60°.

Hope it helps you.....

Answer:

see below

Step-by-step explanation:

      (ab)^n=a^n * b^n

We need to show that it is true for n=1

assuming that it is true for n = k;

(ab)^n=a^n * b^n

( ab) ^1 = a^1 * b^1

ab = a * b

ab = ab

Then we need to show that it is true for n = ( k+1)

or (ab)^(k+1)=a^( k+1) * b^( k+1)

Starting with

  (ab)^k=a^k * b^k    given

Multiply each side by ab

ab *  (ab)^k= ab *a^k * b^k

   ( ab) ^ ( k+1) = a^ ( k+1) b^ (k+1)

Therefore, the rule is true for every natural number n

Hello, n being an integer, we need to prove that one statement depending on n is true, let's note it S(n).

The mathematical induction involves two steps:

Step 1 - We need to prove S(1), meaning that the statement is true for n = 1

Step 2 - for k integer > 1, we assume S(k) and we need to prove that S(k+1) is true.

Imagine that you are a painter and you need to paint all the trees on one side of a road. You have several colours that you can use but you are asked to follow two rules:

Rule 1 - You need to paint the first tree in white.

Rule 2 - If one tree is white you have to paint the next one in white too.

What colour do you think all the trees will be painted?

Do you see why this is very important to prove the two steps as well ?

Let's do it in this example.

Step 1 - for n = 1, let's prove that S(1) is true, meaning  [tex](ab)^1=a\cdot b =a^1\cdot b^1[/tex]

So the statement is true for n = 1

Step 2 - Let's assume that this is true for k, and we have to prove that this is true for k+1

So we assume S(k), meaning that [tex](ab)^k=a^k\cdot b^k[/tex]

and what about S(k+1), meaning [tex](ab)^{k+1}=a^{k+1}\cdot b^{k+1}[/tex] ?

We will use the fact that this is true for k,

[tex](ab)^{k+1}=(ab)\cdot (ab)^k =(ab) \cdot a^k \cdot b^k[/tex]

We can write it because the statement at k is true and then we can conclude.

[tex](ab)^{k+1}=(ab)\cdot (ab)^k =(ab) \cdot a^k \cdot b^k=a^{k+1}\cdot b^{k+1}[/tex]

In conclusion, we have just proved that S(n) is true for any n integer greater or equal to 1, meaning [tex](ab)^{n}=a^{n}\cdot b^{n}[/tex]

Hope this helps.

Do not hesitate if you need further explanation.

Thank you

Answer:

∛18 * ∛18 * 18/(∛18)²

Step-by-step explanation:

Let the surface area of the box be expressed as S = 2(LB+BH+LH) where

L is the length of the box = x

B is the breadth of the box = x

H is the height of the box = h

Substituting this variables into the formula, we will have;

S = 2(x(x)+xh+xh)

S = 2x²+2xh+2xh

S = 2x² + 4xh and the Volume V = x²h

If V = x²h; h = V/x²

Substituting h = V/x² into the surface area will give;

S = 2x² + 4x(V/x²)

Since the volume V = 18cm³

S = 2x² + 4x(18/x²)

S =  2x² + 72/x

Differentiating the function with respect to x to get the minimal point, we will have;

dS/dx = 4x - 72/x²

at dS/dx = 0

4x - 72/x² = 0

- 72/x² = -4x

72 = 4x³

x³ = 72/4

x³  = 18

[tex]x = \sqrt[3]{18}[/tex]

Critical point is at [tex]x = \sqrt[3]{18}[/tex]

If x²h = 18

(∛18)²h =18

h = 18/(∛18)²

Hence the dimension is  ∛18 * ∛18 * 18/(∛18)²

Step-by-step explanation:

f(x) = integral (-8x) dx = -4x^2 + C

f(1) = -3 = -4 + C

C = 1

f(x) = -4x^2 + 1

The particular solution of the differential equation f'(x) = -8x that satisfies the initial condition f(1) = -3 is: f(x) = -4x² + 1.

Here, we have,

To find the particular solution of the differential equation f'(x) = -8x that satisfies the initial condition f(1) = -3,

we can integrate the equation and use the initial condition to determine the constant of integration.

First, integrate both sides of the equation with respect to x:

∫ f'(x) dx = ∫ -8x dx

Integrating, we get:

f(x) = -4x² + C

Now, we can use the initial condition f(1) = -3 to find the value of the constant C.

Substituting x = 1 and f(x) = -3 into the equation, we have:

-3 = -4(1)² + C

-3 = -4 + C

C = -3 + 4

C = 1

Therefore, the particular solution of the differential equation f'(x) = -8x that satisfies the initial condition f(1) = -3 is:

f(x) = -4x² + 1

To learn more on equation click:

brainly.com/question/24169758

#SPJ2

Answer:

6

Step-by-step explanation:

36 - 20 - 32 x 1/4

=> 36 - 20 - 32/4

=> 36 - 20 - 8

=> 36 - 28

=> 6

Answer:

Table in option C represents the linear relationship as the equation, [tex] y = 2x + 6 [/tex]

Step-by-step explanation:

The equation given seems to be wrong. The equation should be [tex] y = 2x + 6 [/tex], because, taking a look at the tables given, the table in option C is the only table that has values that conforms to the equation, [tex] y = 2x + 6 [/tex].

In table C, when x = 2 using the equation, [tex] y = 2x + 6 [/tex], thus,

[tex] y = 2(2) + 6 = 4 + 6 = 10 [/tex].

When x = 3,

[tex] y = 2(3) + 6 = 6 + 6 = 12. [/tex]

Theredore, the equation, [tex] y = 2x + 6 [/tex], represents the relationship between the X and y variables in the table in option C.

Answer:

  1/7^2

Step-by-step explanation:

The applicable rules of exponents are ...

  (a^b)(a^c) = a^(b+c)

  a^-b = 1/a^b

__

Then your expression simplifies to ...

  [tex]7^3\cdot 7^{-5}=7^{3-5}=7^{-2}=\boxed{\dfrac{1}{7^2}}[/tex]

Answer:

The answer is 1/7^2

Step-by-step explanation:

I took the test lol

Answer: There are no real number roots (the two roots are complex or imaginary)

The discriminant D = b^2 - 4ac tells us the nature of the roots for any quadratic in the form ax^2+bx+c = 0

There are three cases

Answer: 0.8749

Step-by-step explanation:

Given, The time, X minutes, taken by Tim to install a satellite dish is assumed to be a normal random variable with mean 127 and standard deviation 20.

Let x be the time taken by Tim to install a satellite dish.

Then, the probability that Tim will takes less than 150 minutes to install a satellite dish.

[tex]P(x<150)=P(\dfrac{x-\text{Mean}}{\text{Standard deviation}}<\dfrac{150-127}{20})\\\\=P(z<1.15)\ \ \ [z=\dfrac{x-\text{Mean}}{\text{Standard deviation}}]\\\\=0.8749\ [\text{By z-table}][/tex]

hence, the required probability is 0.8749.

Answer:

[tex]\boxed{50}[/tex]

Step-by-step explanation:

Because the initial temperature is 40 degrees and it increases by 10, add the two values together to get the final temperature.

40 + 10 = 50

Therefore, the final answer is 50 degrees.

Answer:

50

Step-by-step explanation:

If it starts at 40 degrees and increases 10 degrees, it is going to be 50 degrees.  Increases means adding, so it is asking you to add 10 to 40 which is 50.  If it asks decreases in the future you will have to subtract.

Answer:

25 CAN be written as a fraction.

=> 250/10 = 25

Square root of 14 is 3.74165738677

It is NOT POSSIBLE TO WRITE THIS FULL NUMBER AS A FRACTION,  but if we simplify the decimal like: 3.74, THEN WE CAN WRITE THIS AS A FRACTION

=> 374/100

-1.25 CAN be written as a fraction.

=> -5/4 = -1.25

Square root of 16 CAN also be written as a fraction.

=> sqr root of 16 = 4.

4 can be written as a fraction.

=> 4 = 8/2

Pi = 3.14.........

It is NOT POSSIBLE TO WRITE THE FULL 'PI' AS A FRACTION, but if we simplify 'pi' to just 3.14, THEN WE CAN WRITE IT AS A FRACTION

=> 314/100

.6 CAN be written as a fraction.

=> 6/10 = .6

Answer:

Tom:

initial money = $ 39000

% increased per annum = 2.9%

money gained per annum = 39000 * 2.9/100 = $1131

Allen:

initial money = $ 39000

% increased per annum = 5.7 %

money gained per annum = 39000 * 5.7/100 = $2223

Allen has $ (2223 - 1131) = $ 1192 more than Tom

Answer:

maybe it's 10.because c is 10,b is 10,and so as s.

hence s is 10 also.

Answer:

4, 7, 10, 13

Step-by-step explanation:

Hey there!

Well in the given pattern,

-11, -8, -5, -2, 1

we can conclude that the pattern is +3 every time.

-11 + 3 = -8

-8 + 3 = -5

-5 + 3 = -2

-2 + 3 = 1

And so on

Hope this helps :)

Answer:

Solution : 8i

Step-by-step explanation:

We can use the trivial identities cos(π / 2) = 0, and sin(π / 2) = 1 to solve this problem. Let's substitute,

[tex]8\left[cos\left(\frac{\pi }{2}\right)+isin\left(\frac{\pi \:}{2}\right)\right][/tex] = [tex]8\left(0+1i\right)[/tex]

And of course 1i = i, so we have the expression 8(0 + i ). Distributing the " 8, " 8( 0 ) = 0, and 8(i) = 8i, making the fourth answer the correct solution.

Answer:

the answers are below:

Step-by-step explanation:

a. null hypothesis:

H0: Pw - Pm = 0 (so Pw = Pm)

alternate hypothesis:

H1: Pw - Pm > 0 (so Pw > Pm)

where Pw is the proportion of women

Pm is the proportion of men

b.) proportion of women = o.40

proportion of men =  0.32

sample size of women = 20

sample size of men = 25

[tex]z = 0.4 - 0.32/ \sqrt{((0.4 *0.6)/20) * (0.32 * 0.68)/25)}[/tex]

[tex]z = 0.56[/tex]

c.) p value =

p(z>0.56)

= 0.7123

= 1 - 0.7123

= o.2877 which can be approximated to be 0.288

d. alpha value was set at 0.05

the p value is greater than alpha.

therefore it is not statistically significant.

we conclude that the proportion of roman catholic women is not greater than men.

Answer:

(A) 0.11

(B) 0.0526

(C) Related

(D) 0.28

Step-by-step explanation:

The data provided is:

DC = event that a randomly selected driver is using a cell phone

TA = event that a randomly selected driver has a traffic accident

(A)

From the provided data:

P (DC) = 0.11

(B)

From the provided data:

P (TA) = 0.0526

(C)

To determine whether the events DC and TA are dependent, we need to show that:

[tex]P(DC\cap TA)\neq P(DC)\times P(TA)[/tex]

The value of P (DC ∩ TA) is,

[tex]P(DC\cap TA)=P(DC|TA)\time P(TA)[/tex]

                     [tex]=0.28\times 0.0526\\=0.014728[/tex]

Now compute the value of P (DC) × P (TA) as follows:

[tex]P (DC) \times P (TA)=0.11\times 0.0526=0.005786[/tex]

So, [tex]P(DC\cap TA)\neq P(DC)\times P(TA)[/tex]

Thus, cell phone use while driving and traffic accidents are related.

(D)

The probability that the driver was distracted by a cell phone given that the driver has an accident is:

P (DC | TA) = 0.28

Answer:

  60°

Step-by-step explanation:

You have the rotation matrix ...

  [tex]\left[\begin{array}{cc}\cos{\theta}&-\sin{\theta}\\\sin{\theta}&\cos{\theta}\end{array}\right]=\left[\begin{array}{cc}\dfrac{1}{2}&-\dfrac{\sqrt{3}}{2}\\\dfrac{\sqrt{3}}{2}&\dfrac{1}{2}\end{array}\right][/tex]

This tells you the angle of rotation is ...

  [tex]\tan{\theta}=\dfrac{\sin{\theta}}{\cos{\theta}}=\dfrac{\left(\dfrac{\sqrt{3}}{2}\right)}{\left(\dfrac{1}{2}\right)}=\sqrt{3}\\\\\theta=\arctan{\sqrt{3}}=60^{\circ}[/tex]

The angle of rotation is 60°.

Answer:

B----- 60

Step-by-step explanation:

(1,1) is your answer.

Work is shown below.

Any questions? Feel free to ask.

Answer: (1,1)

Step-by-step explanation:

Answer: 22 quarters

Step-by-step explanation:

Let N be the number of nickels.

Then the number of quarters is (55-N)

The nickels  contribute 5N cents to the total.

The quarters contribute 25*(55-N) cents to the total.

5N + 25*(55-N) = 715

5N + 1375 - 25N = 715

-20N = 715 - 1375 = -660

[tex]N=\frac{-660}{-20}[/tex]

[tex]=33[/tex]

[tex]55-33=22[/tex]

So there is 22 quarters inside the jar.

Check to see if my answer is correct-

33*5 + 22*25 = 715 cents

Answer:

H₀: μ = 93.29 vs. Hₐ: μ ≠ 93.29.

Step-by-step explanation:

In this case we need to test whether the claim made by BC financial aid is true or not.

Claim: The average textbook at BC bookstore costs $93.29.

A null hypothesis is a sort of hypothesis used in statistics that intends that no statistical significance exists in a set of given observations.  

It is a hypothesis of no difference.

It is typically the hypothesis a scientist or experimenter will attempt to refute or discard. It is denoted by H₀.

Whereas, the alternate hypothesis is the contradicting statement to the null hypothesis.

The alternate hypothesis describes direction of the hypothesis test, i.e. if the test is left tailed, right tailed or two tailed.

It is also known as the research hypothesis and is denoted by Hₐ.

The hypothesis to test this claim can be defined as follows:

H₀: The average textbook at BC bookstore costs $93.29, i.e. μ = 93.29.

Hₐ: The average textbook at BC bookstore costs different than $93.29, i.e. μ ≠ 93.29.

Answer:

B = 6

Step-by-step explanation:

2x^2 + 4xy − 48y^2

Factor out 2

2(x^2 + 2xy − 24y^2)

What 2 numbers multiply to -24 and add to 2

-4 *6 = -24

-4+6 = 2

2 ( x+6y)( x-4y)

Answer:

[tex]\huge\boxed{B=6}[/tex]

Step-by-step explanation:

They are two way to solution.

Factor the polynomial on the left side of the equation:

[tex]2x^2+4xy-48y^2=2(x^2+2xy-24y^2)=2(x^2+6xy-4xy-24y^2)\\\\=2\bigg(x(x+6y)-4y(x+6y)\bigg)=2(x+6y)(x-4y)[/tex]

Therefore:

[tex]2x^2+4xy-48y^2=2(x+By)(x-4y)\\\Downarrow\\2(x+6y)(x-4y)=2(x+By)(x-4y)\to\boxed{\bold{B=6}}[/tex]

Multiply everything on the right side of the equation using the distributive property and FOIL:

[tex]2(x+By)(x-4y)=\bigg((2)(x)+(2)(By)\bigg)(x-4y)\\\\=(2x+2By)(x-4y)=(2x)(x)+(2x)(-4y)+(2By)(x)+(2By)(-4y)\\\\=2x^2-8xy+2Bxy-8By^2=2x^2+(2B-8)xy-8By^2[/tex]

Compare polynomials:

[tex]2x^2+4xy-48y^2=2x^2+(2B-8)xy-8By^2[/tex]

From here we have two equations:

[tex]2B-8=4\ \text{and}\ -8B=-48[/tex]

[tex]1)\\2B-8=4[/tex]        add 8 to both sides

[tex]2B=12[/tex]         divide both sides by 2

[tex]B=6[/tex]

[tex]2)\\-8B=-48[/tex]          divide both sides by (-8)

[tex]B=6[/tex]

The results are the same. Therefore B = 6.

Answer:

The answer is 68 6/11

Step-by-step explanation:

If you enter the number into a calculator it shows you the exact decimal, therefore you can identify the answer.

Answer:

It is 68 6/11

Step-by-step explanation:

First I made all of the improper fractions into whole numbers and fractions and just saw which one was in the middle .

Answer:

A

Step-by-step explanation:

we just substitute the value of "a" given in the above expression we get

21-2(3)

21-6=15

Answer:

a. 15

Explanation:

Step 1 - Input the value of 'a' in the expression.

21 - 2a

21 - 2(3)

Step 2 - Multiply two and three

21 - 2(3)

21 - 6

Step 3 - Subtract six from twenty one

21 - 6

15

Therefore, the value of the expression 21 - 2a if a = 3 is a. 15.

The easy thing to do is notice that 1^4 = 1, 2^4 = 16, 3^4 = 81, and so on, so the sequence follows the rule [tex]n^4+1[/tex]. The next number would then be fourth power of 7 plus 1, or 2402.

And the harder way: Denote the n-th term in this sequence by [tex]a_n[/tex], and denote the given sequence by [tex]\{a_n\}_{n\ge1}[/tex].

Let [tex]b_n[/tex] denote the n-th term in the sequence of forward differences of [tex]\{a_n\}[/tex], defined by

[tex]b_n=a_{n+1}-a_n[/tex]

for n ≥ 1. That is, [tex]\{b_n\}[/tex] is the sequence with

[tex]b_1=a_2-a_1=17-2=15[/tex]

[tex]b_2=a_3-a_2=82-17=65[/tex]

[tex]b_3=a_4-a_3=175[/tex]

[tex]b_4=a_5-a_4=369[/tex]

[tex]b_5=a_6-a_5=671[/tex]

and so on.

Next, let [tex]c_n[/tex] denote the n-th term of the differences of [tex]\{b_n\}[/tex], i.e. for n ≥ 1,

[tex]c_n=b_{n+1}-b_n[/tex]

so that

[tex]c_1=b_2-b_1=65-15=50[/tex]

[tex]c_2=110[/tex]

[tex]c_3=194[/tex]

[tex]c_4=302[/tex]

etc.

Again: let [tex]d_n[/tex] denote the n-th difference of [tex]\{c_n\}[/tex]:

[tex]d_n=c_{n+1}-c_n[/tex]

[tex]d_1=c_2-c_1=60[/tex]

[tex]d_2=84[/tex]

[tex]d_3=108[/tex]

etc.

One more time: let [tex]e_n[/tex] denote the n-th difference of [tex]\{d_n\}[/tex]:

[tex]e_n=d_{n+1}-d_n[/tex]

[tex]e_1=d_2-d_1=24[/tex]

[tex]e_2=24[/tex]

etc.

The fact that these last differences are constant is a good sign that [tex]e_n=24[/tex] for all n ≥ 1. Assuming this, we would see that [tex]\{d_n\}[/tex] is an arithmetic sequence given recursively by

[tex]\begin{cases}d_1=60\\d_{n+1}=d_n+24&\text{for }n>1\end{cases}[/tex]

and we can easily find the explicit rule:

[tex]d_2=d_1+24[/tex]

[tex]d_3=d_2+24=d_1+24\cdot2[/tex]

[tex]d_4=d_3+24=d_1+24\cdot3[/tex]

and so on, up to

[tex]d_n=d_1+24(n-1)[/tex]

[tex]d_n=24n+36[/tex]

Use the same strategy to find a closed form for [tex]\{c_n\}[/tex], then for [tex]\{b_n\}[/tex], and finally [tex]\{a_n\}[/tex].

[tex]\begin{cases}c_1=50\\c_{n+1}=c_n+24n+36&\text{for }n>1\end{cases}[/tex]

[tex]c_2=c_1+24\cdot1+36[/tex]

[tex]c_3=c_2+24\cdot2+36=c_1+24(1+2)+36\cdot2[/tex]

[tex]c_4=c_3+24\cdot3+36=c_1+24(1+2+3)+36\cdot3[/tex]

and so on, up to

[tex]c_n=c_1+24(1+2+3+\cdots+(n-1))+36(n-1)[/tex]

Recall the formula for the sum of consecutive integers:

[tex]1+2+3+\cdots+n=\displaystyle\sum_{k=1}^nk=\frac{n(n+1)}2[/tex]

[tex]\implies c_n=c_1+\dfrac{24(n-1)n}2+36(n-1)[/tex]

[tex]\implies c_n=12n^2+24n+14[/tex]

[tex]\begin{cases}b_1=15\\b_{n+1}=b_n+12n^2+24n+14&\text{for }n>1\end{cases}[/tex]

[tex]b_2=b_1+12\cdot1^2+24\cdot1+14[/tex]

[tex]b_3=b_2+12\cdot2^2+24\cdot2+14=b_1+12(1^2+2^2)+24(1+2)+14\cdot2[/tex]

[tex]b_4=b_3+12\cdot3^2+24\cdot3+14=b_1+12(1^2+2^2+3^2)+24(1+2+3)+14\cdot3[/tex]

and so on, up to

[tex]b_n=b_1+12(1^2+2^2+3^2+\cdots+(n-1)^2)+24(1+2+3+\cdots+(n-1))+14(n-1)[/tex]

Recall the formula for the sum of squares of consecutive integers:

[tex]1^2+2^2+3^2+\cdots+n^2=\displaystyle\sum_{k=1}^nk^2=\frac{n(n+1)(2n+1)}6[/tex]

[tex]\implies b_n=15+\dfrac{12(n-1)n(2(n-1)+1)}6+\dfrac{24(n-1)n}2+14(n-1)[/tex]

[tex]\implies b_n=4n^3+6n^2+4n+1[/tex]

[tex]\begin{cases}a_1=2\\a_{n+1}=a_n+4n^3+6n^2+4n+1&\text{for }n>1\end{cases}[/tex]

[tex]a_2=a_1+4\cdot1^3+6\cdot1^2+4\cdot1+1[/tex]

[tex]a_3=a_2+4(1^3+2^3)+6(1^2+2^2)+4(1+2)+1\cdot2[/tex]

[tex]a_4=a_3+4(1^3+2^3+3^3)+6(1^2+2^2+3^2)+4(1+2+3)+1\cdot3[/tex]

[tex]\implies a_n=a_1+4\displaystyle\sum_{k=1}^3k^3+6\sum_{k=1}^3k^2+4\sum_{k=1}^3k+\sum_{k=1}^{n-1}1[/tex]

[tex]\displaystyle\sum_{k=1}^nk^3=\frac{n^2(n+1)^2}4[/tex]

[tex]\implies a_n=2+\dfrac{4(n-1)^2n^2}4+\dfrac{6(n-1)n(2n)}6+\dfrac{4(n-1)n}2+(n-1)[/tex]

[tex]\implies a_n=n^4+1[/tex]