1. problem on binomial

Please solve the problem of Binomial theory.
1) Prove that C0+4C1+8C2+12C3+……..+4nCn=1+n2n-1

2. Originally Posted by somnath6088
Please solve the problem of Binomial theory.
1) Prove that C0+4C1+8C2+12C3+……..+4nCn=1+n2n-1
Are those C's constants or sequences?

3. problem on binomial

Please solve the problem of Binomial theory.
Prove that C0C1+C1C2+C2C3+……………+Cn-1Cn= (2n)!/(n+1)!(n-1)!

4. The C0 implies C(subscript 0).Which are binomial constants.

5. Hello,

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

And the latter term is not understandable...
use parenthesis, power ^, etc...

6. Originally Posted by somnath6088
prove that

nC0 + 4.nC1 + 8.nC2 + 12.nC3 +...........+4n.nCn=1+n2^(n+1)
Since ${n \choose 0} = 1$, you only need to prove the following:

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

Note that:

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

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

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

7. Originally Posted by somnath6088
Please solve the problem of Binomial theory.
1) Prove that C0+4C1+8C2+12C3+……..+4nCn=1+n2n-1

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

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

8. Originally Posted by somnath6088
Please solve the problem of Binomial theory.
Prove that C0C1+C1C2+C2C3+……………+Cn-1Cn= (2n)!/(n+1)!(n-1)!
Translation:

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

Three hints:

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

#2. Consider Vandermonde's Identity.

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

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C0C1 C1C2 . . . Cn-1Cn

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