superpus07 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

Thanks!

The sum of the cubes of first n positive integers is given by

[n(n+1)/2]^2If n = 10, the sum is 55^2.

Now, do you need to learn it up?

No. Even if you didn't know it, you should have tried to look at the pattern.

1^3 = 1 = 1^21^3 + 2^3 = 9 = 3^21^3 + 2^3 + 3^3 = 36 = 6^21^3 + 2^3 + 3^3 + 4^3 = 100 = 10^2You see that when you find the sum of the numbers (not of the cubes) and square the sum, you get the sum of the cubes.

Sum of 1+2+3+4 = 10. Square it to get 100. Sum of

1^3 + 2^3 + 3^3 + 4^3 = 100This is pattern recognition.

You can also use another method - of averaging.

The numbers look like this: 1, 8, 27, 64, 125 ... etc

The numbers on the right 8^3, 9^3, 10^3 will be much larger than those on the left which are small. The average would lie not in the middle but on the right somewhere between 6^3 and 7^3. I would say around 300 so the sum will be around 300*10 = 3000. This leads us to 55^2.

The only hitch might be the 45^2 given which might make you uncomfortable in using this approximation. 45^2 = 2025

To ensure that avg is around 300 and not around 200, notice that from 10^3, you can make five 200s. From 9^3, you can make about four 200s (using some extra). So overall, you can make many more 200s than the required 10. Therefore, the average must be 300 and not 200.

Answer (C).

I think this problem can be quickly solved by just looking at the answer choices.

(A) This can be eliminated since the answer obviously should be greater than 10^3 or 1000

(D) This can be eliminated since the answer should also definitely be less than 10*10^3 or less than 10000

(E) can be eliminated using similar logic for D above.

Now the contenders are 45^2 and 55^2.

Now 45^2 = 2025. and 55^2 = 3025. We know 10^3 = 1000 and 9^3 = 729. So we can easily see that the total of our required sum would exceed 2025. Hence answer = (C).