PLease find the sum of the given series without using calculas---
co + c1/2 + c2/3 + c4/4 ..... +cn/n+1
It is not a doubt --- just found interesting
1NO TAKERS ? All are so dependant on calculas:)
Well, try using calculas and then I think I will post the solution.
66The given sum
S=C_0+\dfrac{C_1}{2}+\dfrac{C_2}{3}+\ldots +\dfrac{C_n}{n+1}
=\dfrac{n!}{0!\ n!}+\dfrac{n!}{2\cdot 1!\ (n-1)!}+\dfrac{n!}{3\cdot 2!\,(n-2)!}+\ldots +\dfrac{n!}{(n+1)\cdot n!\,0!}
=\dfrac{1}{n+1}\left(\dfrac{(n+1)!}{1!\ n!}+\dfrac{(n+1)!}{2!\ (n-1)!}++\ldots +\dfrac{(n+1)!}{(n+1)!\,0!}\right)
=\dfrac{1}{n+1}\left(^{n+1}C_1+\ ^{n+1}C_2+\ldots +\ ^{n+1}C_{n+1}\right)
=\dfrac{2^{n+1}-1}{n+1}
1Okk man i will do in different way....just watch
I will consider the r th term...
Tr = nCr / (r+1)
Sum will be sigma Tr {from r 0 to n} = sigma {0 to n} nCr / (r+1)
------------------------------------------------------= (n+1)/(n+1). nCr /(r+1)
------------------------------------------------------= n+1C r+1/(n+1)
sigma { 0 to n } nCr = 2^n
so sigma (0 to n} nCr = 2^n+1 - 1
so substitute the same and we will get {2^n+1 - 1}/n+1
If you all are able to understand this...i will feel great pleasur
66That's the same thing that I have done in my previous post. (And ppl will be able to understand you better if you write more clearly.)
62For those who want to use calculus,
Integrate (1+x)n=Σ(nCrxr)
(1+x)n+1/(n+1)=Σ(nCrxr+1/(r+1))
Substitute x=1.. hence the answer..