Find the missing side or angle.
Round to the nearest tenth.

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

Answer:

b

Step-by-step explanation:

because its right dummy

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:

33 web pages (at least)

Step-by-step explanation:

We can set up an inequality to represent this, where x represents the number of web pages made.

1.5x > 1.25x + 8

1.5x represents the number of hours it will take normally, and 1.25x + 8 represents the time with the new software. 1.5x (amount of hours using old software) needs to be larger than the time it takes with the new software.

Solve for x:

1.5x > 1.25x + 8

0.25x > 8

x > 32

So, the designer would have to make at least 33 pages.

The number of web pages would the designer have to make in order to save time using the new software will be 33 web pages (at least).

Inequality is the relationship between two expressions that are not equal, employing a sign such as ≠ “not equal to,” > “greater than,” or < “less than.”.

We can set up an inequality to represent this, where x represents the number of web pages made.

1.5x > 1.25x + 8

The time with the new software is represented by 1.25x + 8 and the normal time is represented by 1.5x. The number of hours spent using the old software must be 1.5 times greater than the time spent using the new product.

Solve for x:

1.5x > 1.25x + 8

0.25x > 8

x > 32

Therefore, the number of web pages would the designer have to make in order to save time using the new software will be 33 web pages (at least).

To know more about inequality follow

https://brainly.com/question/24372553

#SPJ2

Answer:

H0: μc ≤ μs    Ha :μc > μs    

Step-by-step explanation:

The null and alternate hypotheses can be stated as

H0: μc ≤ μs    Ha :μc > μs    one tailed test

Where

μc =  Mean of college students watching movies in a month

μs  =  Mean of school students watching movies in a month

For one tailed test of α =0.05   the value of Z= ± 1.645

The critical region will be Z >  ± 1.645

It is of importance to note that by rejecting the null hypothesis and accepting the alternate hypothesis we are automatically rejecting all values of mean that are greater than 7.1

Answer:

Yes because no same x-value resulted in different y-values.

Answer:

Yes

Step-by-step explanation:

Answer:

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

hence s is 10 also.

Answer:

The 90 % confidence limits are (-2.09, 8.09).

Since the calculated values do not lie in the critical region we accept our null hypothesis.

Step-by-step explanation:

The null and alternative hypothesis are given by

H0: σ₁²= σ₂² against Ha: σ₁² ≠ σ₂²

Confidence interval for the population mean difference is given by

(x`1- x`2) ± t √S²(1/n1 + 1/n2)

Where S ²= (n1-1)S₁² + S²₂(n2-1)/n1+n2-2

Critical value of t with n1+n2-2= 50+ 35-2= 83 will be -1.633

Now calculating

S ²=34* (12.8)²+ (14.6)²*49/83= 192.96

Now putting the values in the t- test

(75.1 -72.1) ± 1.633 √ 192.96(1/35 +1/50)

=3 ±  5.09

=-2.09, 8.09  is the 90 % confidence interval for the difference

The 90 % confidence limits are (-2.09, 8.09).

Since the calculated values do not lie in the critical region we accept our null hypothesis.

Answer:

15

Step-by-step explanation:

3 + 3+ 4+ 5+ 5+ 5+ 6+ 7+ 7=45

You would then divide that my 9(the amount of numbers) to get three

(3 + 3+ 4+ 5+ 5+ 5+ 6+ 7+ 7)/9

=3

If you are adding a number the numbers would be

3 + 3+ 4+ 5+ 5+ 5+ 6+ 7+ 7+?/10

Its ten because now you would have 10 numbers.

You know it equals 6, so you ask yourself: What divided by 10 would give you 6 or this equation:

( 3 + 3+ 4+ 5+ 5+ 5+ 6+ 7+ 7+?)/10=6

(45+?)/10=6

multiply both sides of the equal sign by 10

10(45+?)/10=6*10

The 10 on the bottom of the left side cancels out.

(45+?)=60

Subtract 15 from both sides of the equal sign

45+?-45=60-45

?=15

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

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:

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:

Step-by-step explanation:

x² - 5x - 36

To factor the expression rewrite -5x as a difference

That's

x² + 4x - 9x - 36

Factor out x from the expression

x( x + 4) - 9x - 36

Factor out -9 from the expression

x( x + 4) - 9( x+ 4)

Factor out x + 4 from the expression

The final answer is

Hope this helps you

Answer:

[tex] \boxed{(x - 9) \: (x + 4) }[/tex]

Option A is the correct option.-

Step-by-step explanation:

( See the attached picture )

Hope I helped!

Best regards!

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:

x = 1

Step-by-step explanation:

First, move all the variables to one side by subtracting 5x on both sides:

5x + 2 = 12x - 5

2 = 7x - 5

Add 5 to both sides:

7 = 7x

1 = x

Answer:

x=1

Step-by-step explanation:

5x + 2 =12x- 5

Subtract 5x from each side

5x-5x + 2 =12x-5x- 5

2 = 7x-5

Add 5 to each side

2+5 = 7x-5+5

7 = 7x

Divide each side by 7

7/7 = 7x/7

1 =x

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:

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:

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:

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.

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:

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:

Section I test scores are more dispersed that that of section II.

Step-by-step explanation:

Consider the data collected from the arithmetic test given to two sections of a school.

Section I: Mean = 45, Standard  Deviation = 6.5

Section II: Mean = 45,  Standard deviation = 3.1

The mean of both the sections are same, i.e. 45.

So there is no comparison that can be made from the center of the distribution.

The standard  deviation for section I is 6.5 and the standard  deviation for section II is 3.1.

The standard deviation is a measure of dispersion, i.e. it tells us how dispersed the data is from the mean or how much variability is present in the data.

The standard  deviation for section I is higher than that of section II.

So, this implies that section I test scores are more dispersed that that of section II.

Answer:

The equation 1+0=1

Step-by-step explanation:

Other options are not eligible because

1 option -Natural numbers cannot be closed under subtraction

2 option-The equation is not having proper rational numbers, they are decimals

3 option-Whole numbers cannot be closed under subtraction

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:

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: 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.

(1,1) is your answer.

Work is shown below.

Any questions? Feel free to ask.

Answer: (1,1)

Step-by-step explanation:

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]