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Question

Find the summation for the series-: $$5+55+555+5555+...$$

I know it is a duplicate of this, but still, I am posting this because i was thinking of solving another way due to which I got stuck in another series.

My Attempt

$S=5+55+555+5555....$

$2*S=10+110+1110+11110....$

$=10*(1+11+111+1111+...)$

$=\frac{10}{9}*(9+99+999+9999+...)$

$=\frac{10}{9}*((10-1)+(10^{2}-1)+(10^{3}-1)+(10^{4}-1)+...)$

$=\frac{10}{9}*(10*\frac{10^{n}-1}{10-1}-n)$

$=\frac{10}{9}*(\frac{10^{n+1}-10}{9}-n)$

$$S=\frac{5}{9}*(\frac{10^{n+1}-10}{9}-n)$$

So i think i got my answer.But i have doubt in summation of series-:

$S=10+110+1110+11110....$

I got summation (from above ) as-: $=\frac{10}{9}*(\frac{10^{n+1}-10}{9}-n)$

But here , it is different .Which one is correct? I am stuck . Please help me out !

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  • 3
    $\begingroup$ Your answer is correct. The second question you linked to only has answers that give you the $n^{th}$ term of the sequence $10, 110, \ldots$ $\endgroup$ – AnotherJohnDoe Oct 3 '17 at 8:09

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