find the sum of the following series 9+16+29+54+103+ upto n terms - Maths - Sequences and Series

Question

find the sum of the following series 9+16+29+54+103+ upto
n terms

please explain each and every step
THE LINK THAT U HAVE SUGGESTED DOES NOT MATCH
MY QUERY,PLEASE SOLE IT ITS A
DIFFERENT TYPE OF SERIES

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Ajanta Trivedi · Accepted Answer

let $S = 9 + 16 + 29 + 54 + 103 + + t_{n}$ (1) where $t_{n}$ is the nth term of the series $S = 9 + 16 + 29 + 54 + 103 + + t_{n} S = 9 + 16 + 29 + 54 + 103 + + t_{n - 1} + t_{n} 0 = 9 + 7 + 13 + 25 + 49 + (t_{n} - t_{n - 1}) - t_{n}$ [subtracting the corresponding terms ] therefore $t_{n} = 9 + (7 + 13 + 25 + 49 + u p t o (n - 1) t e r m s)$ $t_{n} = 9 + S_{1}$ (2) where $S_{1} = 7 + 13 + 25 + 49 + + t'_{n - 1}$ where $t'_{n - 1}$ is (n-1)th term of S1 (3) $S_{1} = 7 + 13 + 25 + 49 + + t'_{n - 1} S_{1} = 7 + 13 + 25 + + t'_{n - 2} + t'_{n - 1} 0 = 7 + 6 + 12 + 24 + + (t'_{n - 1} - t'_{n - 2}) - t'_{n - 1} t'_{n - 1} = 7 + [6 + 12 + 24 + u p t o (n - 2) t e r m s] t'_{n - 1} = 7 + 6 [1 + 2 + 4 + u p t o (n - 2) t e r m s]$ $t'_{n - 1} = 7 + 6 [1 + 2 + 2^{2} + u p t o (n - 2) t e r m s] t'_{n - 1} = 7 + 6 * \frac{2^{n - 2} - 1}{2 - 1} t'_{n - 1} = 7 + 6 [2^{n - 2} - 1] t'_{n - 1} = 7 + 6 \cdot 2^{n - 2} - 6 t'_{n - 1} = 1 + 6 \cdot 2^{n - 2} t'_{n} = 1 + 6 \cdot 2^{n - 1} [r e p l a c e n - 1 b y n i n w h o l e e q u a t i o n]$ since $S_{1} = \sum_{n = 1}^{n - 1} t'_{n} = \sum_{n = 1}^{n - 1} (1 + 6 \cdot 2^{n - 1})$ $S_{1} = \sum_{n = 1}^{n - 1} 1 + 6 \sum_{n = 1}^{n - 1} 2^{n - 1} = (n - 1) + 6 [1 + 2 + 2^{2} + + 2^{n - 2}] = (n - 1) + 6 * \frac{2^{n - 1} - 1}{2 - 1} = (n - 1) + 6 [2^{n - 1} - 1] = n - 1 + 6 \cdot 2^{n - 1} - 6 S_{1} = n - 7 + 6 \cdot 2^{n - 1}$ therefore $t_{n} = 9 + (n - 7 + 6 \cdot 2^{n - 1}) t_{n} = n + 2 + 6 \cdot 2^{n - 1}$ $S = \sum_{n = 1}^{n} t_{n} = \sum_{n = 1}^{n} (n + 2 + 6 \cdot 2^{n - 1}) = \sum_{n = 1}^{n} n + \sum_{n = 1}^{n} 2 + 6 \sum_{n = 1}^{n} 2^{n - 1} = \frac{n (n + 1)}{2} + 2 n + 6 [1 + 2 + 2^{2} + + 2^{n - 1}]$ $S = \frac{n (n + 1)}{2} + 2 n + 6 * [\frac{2^{n} - 1}{2 - 1}] S = n [\frac{n + 1}{2} + 2] + 6 [2^{n} - 1] S = \frac{n (n + 5)}{2} + 6 \cdot 2^{n} - 6$ hope this helps you cheers!!