Math, asked by sandeepgehlot796, 5 months ago

Find 7 + 77 + 777 + ... up to n terms.​

Answers

Answered by mathdude500
0

Answer:

\boxed{\sf \: 7 + 77 + 777 + 7777 + .. \: n \: terms  =  \: \dfrac{7}{9}\left[\dfrac{10( {10}^{n}  - 1)}{9}  - n \right] \: } \\

Step-by-step explanation:

Given series is

\sf \: 7 + 77 + 777 + 7777 + ... \: n \: terms \\  \\

\sf \:  =  \: 7(1 + 11 + 111 + 1111 + ... \: n \: terms) \\  \\

\sf \:  =  \: \dfrac{7}{9} (9 + 99 + 999 + 9999 + ... \: n \: terms) \\  \\

\sf \:  =  \: \dfrac{7}{9} [(10 - 1) + (100 - 1) + (1000 - 1)  + ... \: n \: terms]\\  \\

\sf \:  =  \: \dfrac{7}{9}[ 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{7}{9}\left[\dfrac{10( {10}^{n}  - 1)}{10 - 1}  - n \times 1 \right] \\  \\

\sf \:  =  \: \dfrac{7}{9}\left[\dfrac{10( {10}^{n}  - 1)}{9}  - n \right] \\  \\

Hence,

\implies\sf \: 7 + 77 + 777 + 7777 + .. \: n \: terms  =  \: \dfrac{7}{9}\left[\dfrac{10( {10}^{n}  - 1)}{9}  - n \right] \\  \\

Similar questions