+1 vote
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Find the infinite sum of the series

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

edited | 166 views
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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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@arvin  how?

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let x= 1+4/7+9/72+16/73+25/74[email protected]

now multiply x by 1/7.

x/7= 1/7 + 4/72 + 9/73+16/74[email protected]

subtracting @1 from @2 we get..

6x/7= 1 + 3/7 +5/72 + 7/73 + 9/74[email protected]

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

6x/49= 1/7 +3/72 + 5/73 + 7/74 [email protected]

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

by Boss (12.2k points)
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@arvin,here if you can see the limit of series is up till infinity. :)
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@devshree yes boss i have considered infinity only.

its 2/7(1+1/7+1/72+1/73+..................................infinity)

since for infinite gp sum is (a/1-(r)) when 0<r<1 and here a=1 and r=1/7

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@arvin,First of all no need to address me as a boss. Brother shall still be fine. :). Yeah!!. I'll take note of what you said. Okay.:)
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@devshree ok ok brother no worriess :p
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@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. :)

+1 vote