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The sum of the following series 5+55+555+… to n terms is

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Solution

$S_{n} = 5 + 55 + 555 + \dots to n terms$
$\begin{array}{l} = \frac{5}{9} (9 + 99 + 999 + \dots to n terms) \\ = \frac{5}{9} [(10 - \underset{}{\underset{}{1}) + (10^{2} - 1) + (10^{3} - 1) + \dots + (10^{n} - 1)]} \\ = \frac{5}{9} [(10 + 10^{2} + \dots to n term) - n] (s u m o f n t e r m s i n G . P = \frac{a (r^{n} - 1)}{r - 1}) \\ = \frac{5}{9} [\frac{10 (10^{n} - 1)}{9} - n] \\ = \frac{5}{81} [10 (10^{n} - 1) - 9 n] \\ = \frac{5}{81} [10^{n + 1} - 9 n - 10] \end{array}$

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Geometric progression

Remember concepts with our Masterclasses.

The sum of the following series 5+55+555+… to n terms is

Sn=5+55+555+… to n terms =59(9+99+999+… to n terms )=5910−1 )+102−1+103−1+…+10n−1 =5910+102+… to n term −n sum of n terms in G.P=arn-1r-1=591010n−19−n=5811010n−1−9n=58110n+1−9n−10

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