Please solve the problem of Binomial theory.
1) Prove that C0+4C1+8C2+12C3+……..+4nCn=1+n2n-1
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.
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
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 $\displaystyle {n \choose k} = {n \choose n - k}$.