For which of the following values of n , the value of 1^2 + 2^2 + 3^2

Question

For which of the following values of n , the value of  1^2 + 2^2 + 3^2 + ... + n^2  is NECESSARILY even?

i. n is even
ii. n is a multiple of 6
iii. n is a multiple of 4

(A) only i
(B) only iii
(C) Both i and ii
(D) Both i and iii 
(E) Both ii and iii

chetan2u · Answer

the value of  1^2 + 2^2 + 3^2 + ... + n^2  as even will be same as for  1+2+3+..+n

Let us work out a pattern
1) 1....Odd
2) 1+2... O+E=Odd
3) 1+2+3...O+E+O=Even
4) 1+2+3+4....O+E+O+E=Even
5) 1+2+3+4+5....O+E+O+E+O=Odd
6) 1+2+3+4+5+6....O+E+O+E+O+E=Odd
and so on

So pattern is O,O,E,E,O,O,E,E....
i. n is even.... When n is 2 or 6..No 
ii. n is a multiple of 6... When n is 6..No or rather whenever n is a multiple of 6, the value will be ODD 
iii. n is a multiple of 4... As can be seen from the pattern, all multiples of 4 will give value as EVEN . Also Multiple of 4 means TWO odd and TWO even, so both will add up to EVEN

B

sujoykrdatta · Answer

Method 1: Check the partern: 
Note that we need an even number of odd terms to add up to an even:

1st term is odd
2nd term is even
3rd term is odd --- 2 odds till here, so even
4th term is even --- 2 odds till here, so even
5th term is odd
6th term is even
7th term is odd --- 4 odds till here, so even
8th term is even --- 4 odds till here, so even

Observe that whenever the number of terms i.e. n is of the form 4k (multiple of 4) or 4k-1, the sum is even

Hence, option B

Method 2: Use summation formula: 
The above summation is n(n+1)(2n+1)/6

Clearly: 
# if n = 4k, though 2 cancels from the 6 in the denominator, the numerator has a 2 remaining => hence the result is even

# if n = 4k-1, the term (n+1) becomes 4k and hence, from the above logic, the result is even

Hence Option B

Posted from my mobile device

ScottTargetTestPrep · Answer

Recall that any power of an odd number is also odd and any power of an even number is also even.

i) If n = 2, then the summation becomes 1^2 + 2^2 = 5, which is odd. Thus, statement i is not necessarily true.

ii) If n = 6, then the summation becomes 1^2 + 2^2 + 3^2 + 4^2 + 5^2 + 6^2; which is odd + even + odd + even + odd + even = odd + odd + odd = odd. Thus, statement ii is not necessarily true.

iii) If n is a multiple of 4, then we can group the terms into groups of 4; thus, each group will consist of two even terms and two odd terms. Since even + even + odd + odd = even + even = even, the summation will always be even. Statement iii is correct.

Answer: B

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For which of the following values of n , the value of 1^2 + 2^2 + 3^2

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For which of the following values of n , the value of 1^2 + 2^2 + 3^2

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