Question

I think it's incorrect !

Answer 1

http://imgur.com/mRAvcns

Answer 2

yeet

Answer 3

^What? @benlindquist

Answer 4

felt like saying that

Answer 5

My Explanation :-
\[\large \bf C_0+C_1x+c_2x^2+C_3x^3+---=(1+x)^n\]
\[\large \bf integerate~w.r.t~x,\]
\[\large \bf C_0x+C_1 \frac{x^2}{2}+C_2\frac{x^3}{3}+----=\frac{(1+x)^{n+1}}{n+1}+ \lambda \]
\[\large \bf Put~x=0,\]
\[\large \bf 0=\frac{1}{n+1}+ \lambda\]
\[\large \bf \lambda=\frac{-1}{n+1}\]

So,\[\large \bf C_0x+C_1 \frac{x^2}{2}+C_2\frac{x^3}{3}+----=\frac{(1+x)^{n+1}}{n+1}- \frac{1}{n+1} \]
\[\large \bf C_0x+C_1 \frac{x^2}{2}+C_2\frac{x^3}{3}+----=\large \bf\frac{(1+x)^{n+1}-1}{n+1}\]
Now, put x=1,
\[\large \bf C_0+\frac{C_1}{2}+\frac{C_2}{3}+----=\frac{2^{n+1}-1}{n+1}\]
\[\large \bf \color{red}{Hence~Proved}\]

Answer 6

@hartnn

Answer 7

@imqwerty

Answer 8

@jigglypuff314

Answer 9

@mathmale

Answer 10

your work is correct! :)

lets verify,
(x+1)^2 = x^2 +2x+1

LHS = 1+ 2/2 + 1/3 = 7/3

RHS = (2^3-1)/(2+1) = 7/3


good :)

Answer 11

yeah thats correct :)

Answer 12

thank you guys ! @hartnn and @imqwerty

Answer 13

So, @imqwerty physicsgalaxy main MATHMANTHAN ek site h, usme yeh error dikha rakha tha !

Answer 14

might be some mistake
if the summation is \(\large \frac{2^{n+1}-1}{n+1}\) then the series must be this->\[\sum_{n=0}^{n-1}\frac{ C_n }{ n+1 }\]

Answer 15

yep