Infinite series

Question

Find the infinite sum of the series

$1 + \frac{4}{7} + \frac{9}{7^2} + \frac{16}{7^3} + \frac{25}{7^4} + .............\Join$

Comments

I guess there's a minor error in ur series. The second term should be 4. See take 1/7 common. It is repeating (1/7)^n times. For the numbers in series above,apply the formula for the summation of squares of natural numbers. I hope this helps. So answer will revolve somewhere around (1/7)^n(Formula for summing the squares of natural numbers)

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@Devshree yea right

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The series following:

$\frac{(1)^{2}}{7^{0}} +\frac{(2)^{2}}{7^{1}} +\frac{(3)^{2}}{7^{2}} +\frac{(4)^{2}}{7^{3}} +\frac{(5)^{2}}{7^{4}} +........+ \frac{(i)^{2}}{7^{i-1}} + ..........$

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Corrected

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answer : 49/27

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@arvin  how?

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answered @pankaj

Answer 1

let x= 1+4/7+9/7²+16/7³+25/7⁴..................................@1

now multiply x by 1/7.

x/7= 1/7 + 4/7² + 9/7³+16/7⁴..................................@2

subtracting @1 from @2 we get..

6x/7= 1 + 3/7 +5/7² + 7/7³ + 9/7⁴............................@3

now multiply @3 by 1/7 we get..

6x/49= 1/7 +3/7² + 5/7³ + 7/7⁴ ..............................@4

subtracting @4 from @3 we get,

36x/49= 1+2/7 + 2/7² + 2/7³ +2/7⁴...............................infinity

36x/49= 1+ 2/7( 1/(1-1/7))

           = 1+ 1/3 =4/3

x= 4/3 *49/36 = 49/27 _answer

Comments

@arvin,Please may I know why did you had to multiply by 1/7?A small doubt here. :)

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@devshree sure , its because we need to simplify at each point in the equation. and if we can however bring the numerator into easy terms we can easily evaluate it as the denominator is in GP so we go by multiplying it by 1/7. and more often its more of an observation on how we can simplify it u can take it as a rule for solving such problems.

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@arvin,Thanks. :)