# 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#

A)

B)

C)

D)

##### 1 Answer

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

# 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