The sum 1 + 3 + 11 + 25 + 45 + 71 + ... upto 20 terms, is equal to
Show Hint
- 7240
- 7130
- 6982
- 8124
The Correct Option is A
Solution and Explanation
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 \( \sum_{n=1}^{20} (3n^2 - 7n + 5) = 3\frac{20 \cdot 21 \cdot 41}{6} - 7\frac{20 \cdot 21}{2} + 5(20) = 7240 \)
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