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What is the sum of the cubes of the first ten positive integ

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by Scott@TargetTestPrep » Sun Jun 23, 2019 10:19 am
BTGmoderatorDC wrote:What is the sum of the cubes of the first ten positive integers?

(A) 10^3
(B) 45^2
(C) 55^2
(D) 100^2
(E) 100^3

OA C

Source: Manhattan Prep
The sum of the cubes of the first n positive integers has the formula [n(n + 1)/2]^2. Thus the sum of the cubes of the first 10 positive integers is [10(11)/2]^2 = [(2 x 5 x 11)/2]^2 = 55^2.

Alternate solution:

Another way to reiterate the above fact is the sum of cubes of the first n positive integers is the square of the sum of the first n positive integers. Thus the sum of the cubes of the first 10 positive integers is

(1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10)^2 = 55^2

(Notice that adding the first 10 positive integers is much faster than actually adding the cubes of the first 10 integers.)

Answer: C

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edited

by deloitte247 » Fri Jun 28, 2019 6:25 am
Sum of cubes of first 2 positive integers
$$1^3+2^3=1+8=9=3^2$$

Sum of the for the first three positive integers.
$$1^3+2^3+3^3=36=6^3$$

The sum of the first four positive integers.
$$1^3+2^3+3^3+4^3=100=10^2$$

The sum of the nth term of =
$$\left[\frac{2n}{4}\left(n+1\right)\right]^2$$

Test for the second term where n=2
$$\left[\frac{\left(2\cdot2\right)}{4}\left(2+1\right)\right]=3^2$$

Test for third term when n=3
$$\left[\frac{\left(2\cdot3\right)}{4}\cdot\left(3+1\right)\right]^2=\left(\frac{24}{4}\right)^2=6^2$$

Test for the fourth term when n=4
$$\left[\frac{\left(2\cdot4\right)}{4}\cdot\left(4+1\right)\right]^2=\left(\frac{40}{4}\right)^2=10^2$$

For the 10th term when n =10
$$\left[\frac{\left(2\cdot10\right)}{4}\cdot\left(10+1\right)\right]^2=\left(\frac{220}{4}\right)^2=55^2$$

$$answer\ is\ Option\ C$$
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