The of the first n terms of the series 1^2 + 2.2^2 + 3^2 + 4.4^2...... is (n(n+1)^2) / 2, where n is even. When n is odd, the sum is?

A) (n^2(n+1)) /2
B) (n(n+1)(2n+1)) /2
C) (n((n+1))^2) /2
D) n^2((n+1))^2 /2

1 Answer
Aug 24, 2017

I have attempted to resolve an ambiguity via comment, but this has not been forthcoming.

Allow this explanation to demonstrate the ambiguity in the sequence:

I am assuming the continuation of the sequence is:

S = 1^2+2.2^2+3^2+4.4^2+5^2+6.6^2 + ... + (n-1)^2 + n.n^2

\ \ = 1^2+2^3+3^2+4^3+5^2+6^3 + ... + (n-1)^2 + n^3

\ \ = {1^2+3^2+5^2 + ... + (n-1)^2 } +
\ \ \ \ \ \ \ \ {2^3+4^3+6^3+ n^3}

ie the terms are the sum of the odd squares and the even cubes .

We are given that when n is even the sum evaluates to:

S_e(n) = 1^2+2^3+3^2+4^3+5^2+6^3 + ... + n^3
\ \ \ \ \ \ \ \ = (n(n+1)^2)/2 ..... [A}

However this summation formula is inconsistent with the sequence as assumed. We can see this as follows:

Let us consider even number of terms:

S(n) = 1^2+2.2^2+3^2+4.4^2+5^2+ ... + (n-1)^2 + n^3

Then by direct evaluation:

S(0) = 0
S(2) = 1^2 + 2^3 = 1+8=9
S(4) = 1^2 + 2^3+3^2+4^3 =9+9+64= 82
S(6) = 1^2 + 2^3+3^2+4^3 +5^2 + 6^3 =82+25+216=323

Let

F(n) = (n(n+1)^2)/2

Then:

F(0) = (0*1^2)/2 = 0
F(2) = (2*3^2)/2 = 18/2 = 9
F(4) = (4*5^2)/2 = 100/2 = 50
F(6) = (6*7^2)/2 = 294/2 = 147

As these results are inconsistent then either the given formula id invalid or the sequence is not as above due the n^(th)' term not clearly being defined.