The $n^{\text {th }}$ term of the series $1+(3+5+7)+(9+11+13+$ $15+17)+\ldots$
The $n^{\text {th }}$ term of the series $1+(3+5+7)+(9+11+13+$ $15+17)+\ldots$
$(2 n+1)\left[n^2-(n-1)^2\right]$
$(2 n-1)\left[(n-1)^2-n^2\right]$
$(2 n+1)\left[(n-1)^2-n^2\right]$
$(2 n-1)\left[(n-1)^2+n^2\right]$
Solution
Given the series
$\underbrace{1}_{t_1}+\underbrace{(3+5+7)}_{t_2}+\underbrace{(9+11+13+15+17)}_{t_3}+\ldots .$
$\because$ Number of numbers in $n^{\text {th }}$ term $=2 n-1$
Let first term in $t_{\mathrm{n}}=a_{\mathrm{n}}$
So, $t_n=\{a, a+2, a+4, \ldots$. upto $(2 n-1)$ terms $\}$
Since, $a_1=1, a_2=3, a_3=9, a_4=19$ as so on, we get
$a_{\mathrm{n}}=2(n-1)^2+1$
Now,
$\begin{aligned}
& \sum t_n=\left\{a_n+a_n+2+a_n+4+\ldots . . \text { upto }(2 n-1) \text { term }\right\} \\
& =\frac{(2 n-1)}{2}\left\{2 \times a_n+(2 n-1-1) \times 2\right\} \\
& =(2 n-1)\left(2(n-1)^2+1+2 n-2\right) \\
& =(2 n-1)\left(2 n^2-2 n+1\right) \\
& \Rightarrow \sum t_n=(2 n-1)\left((n-1)^2+n^2\right)
\end{aligned}$