Alg10.19

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Prove that 2{C \choose 0}+2^2 \frac{{C \choose 1}}{2}+2^3 \frac{{C \choose 2}}{3}+......+2^{n+1}\frac{{C \choose n}}{n+1}=\frac{3^{n+1}-1}{n+1}\,

2{C \choose 0}+2^2 \frac{{C \choose 1}}{2}+2^3 \frac{{C \choose 2}}{3}+......+2^{n+1}\frac{{C \choose n}}{n+1}=2+2^2\frac{n}{2}+2^3\frac{n(n-1)}{3}+......+2^{n-1}\frac{1}{n+1}\,

which is equal to \frac{1}{n+1}[(n+1)2+\frac{n(n+1)}{2!} 2^2+.......+a^{n+1}]\,

\frac{1}{n+1}[1+(n+1)\!C_1\cdot 2+(n+2)\!C_2\cdot 2^2+......+(n+1)\!C_{n+1} 2^{n+1}-1]\,

\frac{1}{n+1}[(1+2)^{n+1}-1]=\frac{3^{n+1}-1}{n+1}\,


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