The sum 1+3+11+25+45+71+.. upto 20 terms, is equal to - Vidyadip

MCQ

The sum $1+3+11+25+45+71+.$. upto 20 terms, is equal to

A
7240
B
7130
C
6982
D
8124

✓

Answer

(A). 7240
Given sum is
$
S_{n}=1+3+11+25+45+71+\ldots+T_{n}
$
First order differences are in A.P.
Thus, we can assume that
$
T_{n}=an^2+bn+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 $a =3, b=-7, c =5$
Hence, general term of given series is
$
T_{n}=3 n^2-7 n+5
$
Hence, required sum equals
$
\sum_{n=1}^{n=20}\left(3 n^2-7 n+5\right)=3\left(\frac{20 \cdot 21 \cdot 41}{6}\right)-7\left(\frac{20 \cdot 21}{2}\right)+5(20)=7240
$

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