If n is a positive integer, is (1^2 + 2^2 + ... + n^2) - (1 + 2 +..+n)

Question

If n is a positive integer, is  (1^2+2^2+…..+n^2) - (1+2+……+n)  divisible by 8?

(1) n is an odd number
(2)  14 < n < 17

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chetan2u · Answer

Sum of squares of first n positive integer =  \frac{n(n+1)(2n+1)}{6} 
Sum of first n positive integer =  \frac{n(n+1)}{2}

(1^2+2^2+…..+n^2) - (1+2+……+n) = \frac{n(n+1)(2n+1)}{6} - \frac{n(n+1)}{2}

\frac{n(n+1)}{2}(\frac{2n+1}{3}-1) = \frac{n(n+1)}{2}(\frac{2n-2}{3})

\frac{n(n+1)}{2}*2(\frac{n-1}{3}) =\frac{(n-1)n(n+1)}{3}

Now, (n-1)n(n+1) is product of three consecutive integers.
a) If n is even, n has to be a multiple of 8 for the entire expression to be multiple of 8.
b) If n is odd, n-1 and n+1 will be even and surely a multiple of 2*4 as one of the two will be a multiple of 4 at least. So, the entire expression will be multiple of 8.

(1) n is an odd number
Case (b) above.
Sufficient

(2)  14 < n < 17 
So, possible values
15: Odd number, so case (b) above. The entire expression will be multiple of 8.
16: Even number and multiple of 8, so case (a) above. The entire expression will be multiple of 8.
Sufficient

D

If n is a positive integer, is (1^2 + 2^2 + ... + n^2) - (1 + 2 +..+n)

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