Determine-the-sum-of-the-1st-nth-term-of-the-Sequence-1-4-10-22-46-Almighty-Formula-

Question Number 120363 by Lordose last updated on 30/Oct/20

Determine the sum of the 1 st nth term

of the Sequence

1, 4, 10, 22, 46, \dots .

★ Almighty Formula

Answered by floor(10²Eta[1]) last updated on 31/Oct/20

(1, 4, 10, 22, 46, \dots, a_{n}) = (a_{1}, a_{2}, a_{3}, \dots, a_{n})

a_{1} = a_{1}

a_{2} = a_{1} + 3

a_{3} = a_{2} + 6

a_{4} = a_{3} + 12

\dots (sum all together)

S_{n} = a_{1} + S_{n - 1} + (3 + 6 + 12 + \dots + b_{n - 1})

S_{n} = a_{1} + S_{n} - a_{n} + (3 + 6 + \dots + b_{n - 1})

a_{n} = a_{1} + (3 + 6 + 12 + \dots + b_{n - 1})

a_{n} = 1 + (3 + 6 + 12 + \dots + b_{n - 1}) [geometric progression]

a_{n} = 1 + 3 (2^{n - 1} - 1)

a_{n} = 3 . 2^{n - 1} - 2

S_{n} = \underset{k = 1}{\sum^{n}} (3 . 2^{k - 1} - 2) = 3 (\underset{k = 1}{\sum^{n}} 2^{k - 1}) - 2 n

\Rightarrow S_{n} = 3 . 2^{n} - 3 - 2 n