Question

The sum of the first n-terms of the series 1²+2.2²+3² +2.4²+5²+2.6²+........... is $\frac{n(n+1)^2}{2}$, when n is even.

When n is odd, the sum is

• A. (n+1)! - 1
• B. (n-1)! - 1
• C. (n-1)! + 1
• D. (n + 1)! + 1

Answer 1

$\frac{n^2(n+1)}{2}$

Answer 2

Let the series be denoted by $S_n$. The $k$-th term of the series, $T_k$, can be defined as: $T_k = k^2$ if $k$ is odd. $T_k = 2k^2$ if $k$ is even.

We are given that the sum of the first $n$ terms, $S_n$, is $\frac{n(n+1)^2}{2}$ when $n$ is an even number.

We need to find the sum $S_n$ when $n$ is an odd number. If $n$ is odd, then $n-1$ is an even number. The sum $S_n$ can be expressed as the sum of the first $n-1$ terms plus the $n$-th term: $S_n = S_{n-1} + T_n$.

Since $n-1$ is even, we can use the given formula for the sum of an even number of terms, by replacing $n$ with $n-1$: $S_{n-1} = \frac{(n-1)((n-1)+1)^2}{2} = \frac{(n-1)(n)^2}{2} = \frac{n^2(n-1)}{2}$.

For an odd value of $n$, the $n$-th term of the series is $T_n = n^2$.

Now, we can substitute these into the equation for $S_n$: $S_n = S_{n-1} + T_n = \frac{n^2(n-1)}{2} + n^2$.

To simplify, we can factor out $n^2$: $S_n = n^2 \left( \frac{n-1}{2} + 1 \right) = n^2 \left( \frac{n-1+2}{2} \right) = n^2 \left( \frac{n+1}{2} \right) = \frac{n^2(n+1)}{2}$.

Comparing this result with the given options for Q.62: (1) $\frac{n(n+1)^2}{4}$ (2) $\frac{n^2(n+2)}{4}$ (3) $\frac{n^2(n+1)}{2}$ (4) $\frac{n(n+2)^2}{4}$

The calculated sum matches option (3).