Question

Find the sum of the n terms of the series 6+66+666+…

Answer 1

Hint: Use the fact that 666…6( n times 6) can be written as $\dfrac{6}{9}\times 999...9=\dfrac{6}{9}\left( {{10}^{n}}-1 \right)$. Hence express all the terms in the given above form. Use the fact that the sum of n terms of a G.P is given by ${{S}_{n}}=\dfrac{a\left( {{r}^{n}}-1 \right)}{r-1}$ and hence find the sum of the series.

Complete step-by-step answer:
We have ${{S}_{n}}=6+66+666+...+\left( upto\ n\ terms \right)$
Now, we know that $666...6\left( n\ times \right)=\dfrac{6}{9}\left( 999...9 \right)=\dfrac{6}{9}\left( {{10}^{n}}-1 \right)$
Hence, we have
${{S}_{n}}=\dfrac{6}{9}\left( {{10}^{1}}-1 \right)+\dfrac{6}{9}\left( {{10}^{2}}-1 \right)+...+\dfrac{6}{9}\left( {{10}^{n}}-1 \right)$
Taking $\dfrac{6}{9}$ common from the terms of the series, we get
${{S}_{n}}=\dfrac{6}{9}\left( 10-1+{{10}^{2}}-1+...+{{10}^{n}}-1 \right)$
Hence, we have
${{S}_{n}}=\dfrac{6}{9}\left( 10+{{10}^{2}}+{{10}^{3}}+...+{{10}^{n}}-n \right)$
Now the series $10+{{10}^{2}}+...+{{10}^{n}}$ is the sum up to n terms of the geometric progression $10,{{10}^{2}},{{10}^{3}},...$
Hence, we have
${{S}_{n}}=\dfrac{6}{9}\left( \dfrac{10\left( {{10}^{n}}-1 \right)}{10-1}-n \right)=\dfrac{6}{9}\left( \dfrac{10}{9}\left( {{10}^{n}}-1 \right)-n \right)$, which is the required sum up to n terms of the given series.

Note: Verification:
We know that ${{S}_{n}}-{{S}_{n-1}}={{a}_{n}}$
Now, we have
${{S}_{n}}-{{S}_{n-1}}=\dfrac{6}{9}\left( \dfrac{10}{9}\left( {{10}^{n}}-1-{{10}^{n-1}}+1 \right)-\left( n \right)+n-1 \right)=\dfrac{6}{9}\left( \dfrac{10}{9}\left( {{10}^{n-1}} \right)\times 9-1 \right)=\dfrac{6}{9}\left( {{10}^{n}}-1 \right)$
Also the nth term of the given series $=\dfrac{6}{9}\left( {{10}^{n}}-1 \right)$
Hence, we have
${{S}_{n}}-{{S}_{n-1}}={{a}_{n}}$
Hence the answer is verified to be correct.