Question

Sum of infinity of series 1+4/5+7/5^2+10/5^3+.... is

Answer 1

Answer is 35/16

Its correct explanation is below

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Answer 2

Answer:

$\frac{35}{16}$ is the sum of infinity of series $1 + \frac{4}{5} + \frac{7}{5^2} + \frac{10}{5^3} + ......$

Step-by-step explanation:

Explanation:

Given, a infinite series $1 + \frac{4}{5} + \frac{7}{5^2} + \frac{10}{5^3} + ......$

Let S =( $1 + \frac{4}{5} + \frac{7}{5^2} + \frac{10}{5^3} + ......$ ) ..........(i)

Now, we multiply  both the side by $\frac{1}{5}$,

$\frac{S}{5} = (\frac{1}{5} + \frac{4}{5^2} + \frac{7}{5^3} +\frac{10}{5^4} +..........$) .....(ii)

Step 1:

On subtracting (i) and (ii) we get,

S - $\frac{S}{5}$ = $(1 + \frac{4}{5} + \frac{7}{5^2} + \frac{10}{5^3} + ......)$ - $(\frac{1}{5} + \frac{4}{5^2} + \frac{7}{5^3} +\frac{10}{5^4} +..........)$

⇒$\frac{5S - S}{5}$ = $1 + (\frac{4-1}{5} + \frac{7-4 }{5^2} +\frac{10 - 7}{5^3} + .......... )$

⇒$\frac{4S}{5}$ = $1 +(\frac{3}{5} + \frac{3}{5^2} + \frac{3}{5^3} + ............. )$

⇒$\frac{4S}{5} = 1 + \frac{3}{5}(1 + \frac{1}{5} + \frac{1}{5^2}+ ....... )$ .........(iii)

Now, as we know that, sum of infinite GP = $\frac{a}{(1 - r)}$

Where r is the common ratio and a is the first term,

We have , $(1 + \frac{1}{5} + \frac{1}{5^2}+ ....... )$

a = 1 and common ratio = $\frac{1}{5}$

$S_n$ = $\frac{(1 )}{1-\frac{1}{5} }$ = $\frac{5}{4}$  , now put this value in equation (iii) we get,

$\frac{4S}{5} = 1 + \frac{3}{5}(\frac{5}{4} )$

⇒$\frac{4S}{5} = 1 + \frac{3}{4}$ ⇒$\frac{4S}{5} = \frac{4 + 3}{4}$

⇒S = $\frac{7}{4} . \frac{5}{4}$ = $\frac{35}{16}$

Final answer:

Hence, the sum of infinity of the given series is $\frac{35}{16}$ .

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