Question

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`

Answer 1

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.