Math, asked by ravisubramanian98, 4 months ago

Find the sum of n terms of series 6+66+666+......n terms ​

Answers

Answered by mathdude500
3

Answer:

\boxed{\sf \: 6 + 66 + 666 + 6666 + .. \: n \: terms  =  \: \dfrac{2}{3}\left[\dfrac{10( {10}^{n}  - 1)}{9}  - n \right] \: } \\

Step-by-step explanation:

Given series is

\sf \: 6 + 66 + 666 + 6666 + ... \: n \: terms \\  \\

\sf \:  =  \: 6(1 + 11 + 111 + 1111 + ... \: n \: terms) \\  \\

\sf \:  =  \: \dfrac{6}{9} (9 + 99 + 999 + 9999 + ... \: n \: terms) \\  \\

\sf \:  =  \: \dfrac{2}{3} [(10 - 1) + (100 - 1) + (1000 - 1)  + ... \: n \: terms]\\  \\

\sf \:  =  \: \dfrac{2}{3}[ 10 + 100 + 1000 + .. \: n \: terms) - (1 + 1 + 1 + .. \: n \: terms)] \\  \\

We know,

Sum of n terms of a GP series having first term a and common ratio r is given by

\begin{gathered}\begin{gathered}\bf\: S_n = \begin{cases} &\sf{\dfrac{a( {r}^{n} - 1) }{r - 1} }, \:  \: r \:  \ne \: 1 \\ \\  &\sf{\qquad \: na, \:   \:  \: \: r = 1} \end{cases}\end{gathered}\end{gathered} \\  \\

So, using these results, we get

\sf \:  =  \: \dfrac{2}{3}\left[\dfrac{10( {10}^{n}  - 1)}{10 - 1}  - n \times 1 \right] \\  \\

\sf \:  =  \: \dfrac{2}{3}\left[\dfrac{10( {10}^{n}  - 1)}{9}  - n \right] \\  \\

Hence,

\implies\sf \: 6 + 66 + 666 + 6666 + .. \: n \: terms  =  \: \dfrac{2}{3}\left[\dfrac{10( {10}^{n}  - 1)}{9}  - n \right] \\  \\

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