The sum $1+3+11+25+45+71+.$. upto 20 terms, is equal to
- $7240$
- $7130$
- $6982$
- $8124$
Solution
$S_n=1+3+11+25+45+71+\ldots+T_n$
First order differences are in A.P.
Thus, we can assume that
$\mathrm{T}_{\mathrm{n}}=\mathrm{an}^2+\mathrm{bn}+\mathrm{c}$
Solving $\left\{\begin{array}{c}T_1=1=a+b+c \\ T_2=3=4 a+2 b+c \\ T_3=11=9 a+3 b+c\end{array}\right\}$,
we get $\mathrm{a}=3, \mathrm{~b}=-7, \mathrm{c}=5$
Hence, general term of given series is
$\mathrm{T}_{\mathrm{n}}=3 \mathrm{n}^2-7 \mathrm{n}+5$
Hence, required sum equals
$\sum_{\mathrm{n}=1}^{\mathrm{n}=20}\left(3 \mathrm{n}^2-7 \mathrm{n}+5\right)=3\left(\frac{20 \cdot 21 \cdot 41}{6}\right)-7\left(\frac{20 \cdot 21}{2}\right)+5(20)=7240$ *
Asked in: JEE Main 2025 (03 Apr Shift 1)