1plz somebody answer
another one
sum to 'n' terms :
11+x + 21+x2 + 41+x4 +............+ 2nx2n+1
plzzzzz reply
Sum to 'n' terms
1(1 - x)(1 - x3) + x2(1 - x3)(1 - x5) + x4(1 - x5)(1 - x7) + ..........
also find Sum if 'n' = ∞ , given that |x|<1
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9 Answers
1111/ (1-x)(1-x3) can be written as 1/x2(x-1)[(1-x) - (1-x3)/ (1-x)(1-x3)]....
the other terms can also be written in this manner....
therefore the sum becomes
1/x2(x-1) [ 1/(1-x3) - 1/(1-x) + 1/(1-x5) - 1/(1-x3)................+1/(1-xn)]
as n→∞ therefore xn =0 when mod x less than 1
=1/x2(x-1) [ 1-1/(1-x) ]
=1/x(1-x)2....
is this the ans...???
23S=log(1+x)(1+x2)(1+x4)....(1+x2n)
multiply divide by (1-x) inside log
so S = log ( 1-x4n1-x)..........(1)
also S=log(1+x)+log(1+x2)+...log(1+x2n)
now check the derivative of S at x = 1
u get ur ans
so just differentiate eq (1) and put x =1
1@joydoot
i can't understand what u have done in first step it is too messy
and ans is a bit different from this
@qwerty
for second the ans is in terms of both x and n ...plz explain in a detailed way
i don't get what u are saying
derivative of S=log(1+x)+log(1+x2)+...log(1+x2n) will give
11+x + 2x1+x2 ...... 2nx2n-11+x2n
but the series is - 11+x + 21+x2 ......+\frac{2^{n}}{x^{2^{n}}+1}
in denominator it is x to power 2 to power 'n' and not x to power 2n ,i made mistake in writing the question before
plz correct me if im wrong but im not able to understand ur solution
341Let S = \frac{1}{1+x} + \frac{2}{1+x^2}+...+\frac{2^n}{1+x^{2^n}}
What is S + \frac{1}{1-x} ?