The sum 1 + 3 + 11 + 25 + 45 + 71 + ... up to20 terms, is equal to [2025]

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Solution

Q.
The sum 1 + 3 + 11 + 25 + 45 + 71 + ... up to 20 terms, is equal to [2025]

1 7240

2 8124

3 7130

4 6982

Ans.
(1)

Let $S_{n}$ = 1 + 3 + 11 + 25 + 45 + 71 + ... + $T_{n}$

Here, series of differences i.e., (3 – 1), (11 – 3), (25 – 11) ..... i.e., 2, 8, 14, .... is in A.P.

If the second order differences of a square are in A.P. then general term is given by

$T_{n} = a n^{2} + b n + c$

Put n = 1, 2, 3, we get

$T_{1}$ = 1 = a + b + c ... (i)

$T_{2}$ = 3 = 4a + 2b + c ... (ii)

$T_{3}$ = 11 = 9a + 3b + c ... (iii)

On solving equation (i), (ii) and (iii), we get the general term of given series as

$T_{n} = 3 n^{2} - 7 n + 5$

Hence, $\sum_{n = 1}^{20} (3 n^{2} - 7 n + 5)$

$= 3 (\frac{20 \cdot 21 \cdot 41}{6}) - 7 (\frac{20 \cdot 21}{2}) + 5 (20) = 7240$ .

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