# Find the General formula for the sum of a series

# Find the sum of the series 5 + 55 + 555 +… up to *n* terms

Find the general formula for sum of the first *n* terms of the series 5 + 55 + 555 +5555 + ……

## Solution: sum of a series

\begin{aligned} \mathrm{Let} \ \ &\mathrm{S}_n = 5+55+555+.....\mathrm{up \ to} \ {\it{n}}\\ \\ &\Rightarrow \ \mathrm{S}_n=5(1+11+111+.....\mathrm{up \ to} \ {\it{n}}) \\ \\ &\Rightarrow \ \mathrm{S}_n=\dfrac{5}{9}\times (9+99+999+.....\mathrm{up \ to} \ {\it{n}}) \\ \\ &\Rightarrow \ \mathrm{S}_n=\dfrac{5}{9}\times (10-1+100-1+1000-1+.....\mathrm{up \ to} \ {\it{n}}) \\ \\ &\Rightarrow \ \mathrm{S}_n=\dfrac{5}{9}\times (10+100+1000+.....\mathrm{up \ to} \ {\it{n}}) \\ & \\ & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ -\dfrac{5}{9}\times (1+1+1+.....\mathrm{up \ to} \ {\it{n}}) \\ \\ &\Rightarrow \ \mathrm{S}_n=\dfrac{5}{9}\times (10+10^2+10^3+.....\mathrm{up \ to} \ {\it{n}}) -\dfrac{5n}{9} \end{aligned}

## 10 + 10^{2} + 10^{3} +….. up to *n* is in geometric progression

We Know

\begin{aligned} &\mathrm{Sum \ of \ a \ geometric \ progression=}\dfrac{a(r^n-1)}{r-1}\\ \\ &Where \ \ \ \ \ \ a=\mathrm{First \ term} \ \ and \ \ r =\mathrm{Common \ ratio} \\ \end{aligned}

Now

\begin{aligned} &\ \mathrm{S}_n=\dfrac{5}{9}\times \bigg(\dfrac{10(10^n-1)}{10-1}\bigg) -\dfrac{5n}{9}\\ \\ &\Rightarrow \ \mathrm{S}_n=\dfrac{5}{9}\times \bigg(\dfrac{10(10^n-1)}{9}-n\bigg) \\ \\ &\Rightarrow\ \mathrm{S}_n=\dfrac{5}{81}\times (10^{n+1}-10-9n) \\ \\ &\Rightarrow\ \mathrm{S}_n=\dfrac{5(10^{n+1}-9n-10)}{81} \\ \\ \end{aligned}

That is, **5 + 55 + 555 +… up to n = 5(10^{n+1} − 9n − 10)/81**