Question
The sum of the first $$n$$ terms of the series $${1^2} + {2.2^2} + {3^2} + {2.4^2} + {5^2} + {2.6^2} + ....{\text{ is }}\frac{{n{{\left( {n + 1} \right)}^2}}}{2}$$ when $$n$$ is even. When $$n$$ is odd the sum is
A.
$${\left[ {\frac{{n\left( {n + 1} \right)}}{2}} \right]^2}$$
B.
$$\frac{{{n^2}\left( {n + 1} \right)}}{2}$$
C.
$$\frac{{n{{\left( {n + 1} \right)}^2}}}{4}$$
D.
$$\frac{{3n\left( {n + 1} \right)}}{2}$$
Answer :
$$\frac{{{n^2}\left( {n + 1} \right)}}{2}$$
Solution :
If $$n$$ is odd, the required sum is
$$\eqalign{
& {1^2} + {2.2^2} + {3^2} + {2.4^2} + ...... + 2.{\left( {n + 1} \right)^2} + {n^2} \cr
& = \frac{{\left( {n - 1} \right){{\left( {n - 1 + 1} \right)}^2}}}{2} + {n^2} \cr
& \left[ {\because \left( {n - 1} \right){\text{ is even}}} \right. \cr
& \therefore {\text{ using given formula for the sum of }}\left( {n - 1} \right)\left. {{\text{terms}}{\text{.}}} \right] \cr
& = \left( {\frac{{n - 1}}{2} + 1} \right){n^2} \cr
& = \frac{{{n^2}\left( {n + 1} \right)}}{2} \cr} $$