Quote:
Originally Posted by poojadatta
49/27
S = 1+4/7+(9/7^2)+(16/7^3)+(25/7^4)+......
7S = 7+4+(9/7)+(16/7^2)+(25/7^3)+......
6S = 10+(5/7)+(7/7^2)+(9/7^3)+......
7*6S = 70+5+(7/7)+(9/7^2)+(11/7^3)......
36S = 65 + 2(1/7 + 1/7^2 + .......)
= 65 + 2*1/6 = 196/3
=> S = 49/27
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Ok this might be a silly question but I did not get the bold part.. can some help[/quote]
@pooja no question is silly to ask..
i tell you a standard method to tackle these type of problems..
whenever there is a series...
which gives a hint of GP and AP .. we always do sum by
multiplying by the common ratio...
here by observation we can find that CR is 1/7..
and the double difference of numerators are in AP...
so s=1+4/7+9/7^2+16/7^3...
1/7s= 1/7 + 4/7^2 + 9/7^3..
on subtracting : 6/7s= 1+ 3/7+5/7^2+..
now numerators are in AP.... so again repeat the same operation..
6/49s= 1/7+ 3/7^2+...
so 6/7s-6/49s= 1+ 2/7+2/7^2+....
i think u can proceed from here..