Find the sum to n terms of the sequence,8, 88 | vedclass

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(B) The given sequence is $8, 88, 888, 8888, \ldots$ to $n$ terms.Let $S_n = 8 + 88 + 888 + 8888 + \ldots$ to $n$ terms.$S_n = 8(1 + 11 + 111 + 1111 +

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$\frac{80}{81}(10^{n}-1)-\frac{8}{9}n$

Vedclass · Answer

(B) The given sequence is $8, 88, 888, 8888, \ldots$ to $n$ terms.Let $S_n = 8 + 88 + 888 + 8888 + \ldots$ to $n$ terms.$S_n = 8(1 + 11 + 111 + 1111 + \ldots 	ext{ to } n 	ext{ terms})$.$S_n = \frac{8}{9}(9 + 99 + 999 + 9999 + \ldots 	ext{ to } n 	ext{ terms})$.$S_n = \frac{8}{9}[(10-1) + (10^2-1) + (10^3-1) + \ldots + (10^n-1)]$.$S_n = \frac{8}{9}[(10 + 10^2 + 10^3 + \ldots + 10^n) - (1 + 1 + 1 + \ldots + 1 	ext{ to } n 	ext{ terms})]$.The sum of the geometric series $10 + 10^2 + \ldots + 10^n$ is $\frac{10(10^n-1)}{10-1} = \frac{10}{9}(10^n-1)$.Thus,$S_n = \frac{8}{9} [\frac{10}{9}(10^n-1) - n]$.$S_n = \frac{80}{81}(10^n-1) - \frac{8}{9}n$.

Find the sum to $n$ terms of the sequence,$8, 88, 888, 8888, \ldots$

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