What is the sum of the cubes of the first ten positive integers?
The sum of the cubes of first n positive integers is given by [n(n+1)/2]^2
If 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 = 10^2
You 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 = 100
This 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.
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).