The \(n\)th term of the series \(1+3+7+13+21+\ldots \ldots\) is 9901 . The value of \(\mathrm{n}\) is

Given, series
\(1+3+7+13+21+\ldots \ldots\)
Also, \(\quad \mathrm{t}_{\mathrm{n}}=9901 \quad \text{...(i)}\)
Let \(\quad \mathrm{S}_{\mathrm{n}}=1+3+7+13+21+\ldots \ldots \mathrm{n}\)
and \(S_{n}=1+3+7+13+\ldots\) n terms
On subtracting
\(0=(1+2+4+6+8+\ldots . .)-t_{n}\)
\(\mathrm{t}_{\mathrm{n}}=1+2+4+6+8+\ldots \ldots \mathrm{n}\) terms
\(\mathrm{t}_{\mathrm{n}}=1+2[1+2+3+4+\ldots \ldots(\mathrm{n}-1)\) terms \(]\)
\(\mathrm{t}_{\mathrm{n}}=1+2\left[\frac{(\mathrm{n}-1)(\mathrm{n}-1+1)}{2}\right]\)
\(\mathrm{t}_{\mathrm{n}}=1+\mathrm{n}(\mathrm{n}-1)\)
\(9901=1+n(n-1)\) [from Eq. (i)
\(\mathrm{n}^{2}-\mathrm{n}-9900=0\)
\(n^{2}-100 n+99 n-9900=0\)
\[
\begin{gathered}
\mathrm{n}(\mathrm{n}-100)+99(\mathrm{n}-100)=0 \\
(\mathrm{n}-100)(\mathrm{n}+99)=0 \\
\mathrm{n}=100 \quad(\mathrm{n}=-99, \text { neglecting })
\end{gathered}
\]
(because terms not negative)