Question

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

• A. 7240
• B. 7130
• C. 6982
• D. 8124

Hint:

If the first differences of a series form an arithmetic progression (AP), then the general term of the series can be represented by a quadratic equation \( T_n = an^2 + bn + c \).

Answer 1

The Correct Option is A

Given sum is \( S_n = 1 + 3 + 11 + 25 + 45 + 71 + ... + T_n \) 
First order differences are in A.P. 
Thus, we can assume that \( T_n = an^2 + bn + c \) 
Solving \( \begin{cases} T_1 = 1 = a + b + c T_2 = 3 = 4a + 2b + c \\T_3 = 11 = 9a + 3b + c \end{cases} \) 
we get a = 3, b = -7, c = 5 
Hence, general term of given series is \( T_n = 3n^2 - 7n + 5 \) 
Hence, required sum equals \( \sum_{n=1}^{20} (3n^2 - 7n + 5) = 3\frac{20 \cdot 21 \cdot 41}{6} - 7\frac{20 \cdot 21}{2} + 5(20) = 7240 \)