Question

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Q. Find Sum to n terms.


8 + 88 + 888 + 8888/........n terms.

Answer 1

${\bold{\underline{Hey\:Mate!}}}$

Here's the Answer ❗

${\boxed{\mathcal{Divergent \:Series}}}$

8 + 88 + 888 + 8888.......n terms.

The series given is neither in G.P nor in A.P it is a divergent series that doesn't have a numerical answer . So firstly we convert it into a geometric progression.

The series can be written as

8 ( 1 + 11 + 111.........n terms )

Dividing by 9 →

$= > \frac{8}{9} \: (9 + 99 + 999.......n \: terms) \\ \\ = > \frac{8}{9} \: ( (10 - 1)(10 {}^{2} - 1)(10 {}^{3} - 1)....n \: terms)$

$= > \frac{8}{9} \: \: (10 + 10 {}^{2} + 10 {}^{3} ....) \: - \: (1 + 1 + 1....)$

Now here ,

10 , 10² , 10³ are in Geometric Progression with both first term and common ratio being equal to 10.

$= > \frac{8}{9} \times( 10 \: \: ( \frac{10 {}^{n} - 1}{9} ) - n )\: \\ \\ = > \frac{80}{81} \: (10 {}^{n} - 1) - \frac{8}{9} n$

This is the final Sum of the series up to n terms.

${\bold{Note:}}$

Formula used here is for Sum of n terms in G.P

When r > 1 then we have

Sn = $\frac{a(r^n- 1)}{r - 1}$

${\red{\huge{\mathtt{Hope\:Helps!!}}}}$

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Answer 2

$= > \frac{8}{9} \: (9 + 99 + 999.......n \: terms) \\ \\ = > \frac{8}{9} \: ( (10 - 1)(10 {}^{2} - 1)(10 {}^{3} - 1)....n \: terms)$

$= > \frac{8}{9} \: \: (10 + 10 {}^{2} + 10 {}^{3} ....) \: - \: (1 + 1 + 1....)$

$= > \frac{8}{9} \times( 10 \: \: ( \frac{10 {}^{n} - 1}{9} ) - n )\: \\ \\ = > \frac{80}{81} \: (10 {}^{n} - 1) - \frac{8}{9} n$

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