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 \([n(n+1)/2]^2\)
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.
\(1^3 = 1 = 1^2\)
\(1^3 + 2^3 = 9 = 3^2\)
\(1^3 + 2^3 + 3^3 = 36 = 6^2\)
\(1^3 + 2^3 + 3^3 + 4^3 = 100 = 10^2\)
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.
Answer (C).
_________________
Karishma
Veritas Prep GMAT Instructor
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