Please solve the problem of Binomial theory.

1) Prove that C0+4C1+8C2+12C3+……..+4nCn=1+n2n-1

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- Jun 17th 2008, 05:22 AMsomnath6088problem on binomial
Please solve the problem of Binomial theory.

1) Prove that C0+4C1+8C2+12C3+……..+4nCn=1+n2n-1 - Jun 17th 2008, 05:23 AMcolby2152
- Jun 17th 2008, 05:27 AMsomnath6088problem on binomial
Please solve the problem of Binomial theory.

Prove that C0C1+C1C2+C2C3+……………+Cn-1Cn= (2n)!/(n+1)!(n-1)! - Jun 17th 2008, 05:32 AMsomnath6088
The C0 implies C(subscript 0).Which are binomial constants.

- Jun 17th 2008, 09:44 AMMoo
Hello,

Please use (n,p) if you mean pCn, like you wrote...

And the latter term is not understandable...

use parenthesis, power ^, etc... - Jun 18th 2008, 12:49 AMmr fantastic
Since $\displaystyle {n \choose 0} = 1$, you only need to prove the following:

$\displaystyle 4 \, {n \choose 1} + 8 \, {n \choose 2} + 12 \, {n \choose 3} + ..... + 4n \, {n \choose n} = n 2^{n+1}$

Note that:

$\displaystyle (1 + x)^n = \sum_{k=0}^{n} {n \choose k} \, x^k$ .... (1)

Differentiate both sides of (1) with respect to x:

$\displaystyle n (1 + x)^{n-1} = \sum_{k=0}^{n} {n \choose k} \, k\, x^k$ .... (2)

(Identity (2) can also be proved using a combinatorial argument)

Now substitute x = 1 into (2). Then multiply both sides by 4 and you get the result. - Jun 18th 2008, 02:13 AMmr fantastic
Is it meant to read

$\displaystyle {n \choose 0} + 4 \, {n \choose 1} + 8 \, {n \choose 2} + 12 \, {n \choose 3} + ..... + 4n \, {n \choose n} = 1 + n 2^{n-1}$ ?

This statement is FALSE. Substitute n = 1, for example, to see this.

The correct identity is given (and a solution suggested) in this thread: http://www.mathhelpforum.com/math-he...l-problem.html - Jun 18th 2008, 03:01 AMmr fantastic
Translation:

Prove that $\displaystyle {n \choose 0} \, {n \choose 1} + {n \choose 1} \, {n \choose 2} + {n \choose 2} \, {n \choose 3} + \, ..... \, + {n \choose n-1} \, {n \choose n} = \frac{(2n)!}{(n+1)! \, (n - 1)!}$.

Three hints:

#1. Note that the right hand side is equivalent to $\displaystyle {2n \choose n-1}$. This suggests hint #2.

#2. Consider Vandermonde's Identity.

#3. Note that $\displaystyle {n \choose k} = {n \choose n - k}$.