Question

Find the sum of the series : 0.7 +0.77 +0.777 +............upto n terms.

Answer 1

Answer: The sum would be $\dfrac{7}{81}[9n-1+10^{-n}]$bvcbcbc

Step-by-step explanation:

Since we have given that

$0.7+0.77+0.777+.......................$

We can rewrite it as :

$\dfrac{7}{10}+\dfrac{7}{100}+\dfrac{777}{1000}+..............$

Insert 9 and take out 7:

$\dfrac{7}{9}(0.9+0.99+0.999+..............)\\\\=\dfrac{7}{9}(\dfrac{9}{10}+\dfrac{99}{100}+\dfrac{999}{1000}+.................)$

So, it becomes

$\dfrac{7}{9}[(1-0.1)+(1-0.01)+(1-0.001)+................]\\\\=\dfrac{7}{9}(1+1+1+...................)-(0.1+0.01+0.001+..................)\\\\=\dfrac{7}{9}(n-\dfrac{0.1(1-0.1^n}{1-0.1})\\\\=\dfrac{7}{81}[9n-1+10^{-n}]$

Hence, the sum would be $\dfrac{7}{81}[9n-1+10^{-n}]$

Answer 2

$\huge \bold \color{lime} \star {\underline \color{deeppink} \mathcal{HEYA}} \star$

$0.7 + 0.77 + 0.777 \: .......n \: terms$

$\frac{7}{10} + \frac{77}{100} + \frac{777}{1000} .......n \: terms$

Taking 7 common and multiplying and dividing by 9, we get

$\frac{7}{9} ( \frac{9}{10} + \frac{99}{100} + \frac{999}{1000 } + .......)$

$\frac{7}{9} (0.9 + 0.99 + 0.999 + .....)$

$\small{\frac{7}{9} ((1 - 0.1) + (1 - 0.01) + (1 - 0.001) + .........)}$

$\small {\frac{7}{9} ((1 + 1 + 1 + ....) - (0.1 + 0.01 + 0.001 + ......))}$

$\frac{7}{9} (n - \frac{0.1(1 - {0.1}^{n}) }{1 - 0.1} )$

$\frac{7}{9} (n - \frac{0.1(1 - {0.1}^{n} )}{0.9} )$

$\frac{7}{81} (9n - 1 + {0.1}^{n} )$

$\frac{7}{81} (9n - 1 - \frac{1}{ {10}^{n} } )$

$\bold \color{green} \frac{7}{81} (9n - 1 + {10}^{ - n} )$

Hope it helps ......

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