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Prove that C0 +2C1 + 3C2 +....... + (n+1)Cn = 2^(n-1) × (n+2)

Prove that C0 +2C1 + 3C2 +....... + (n+1)Cn = 2^(n-1) × (n+2)

Grade:12th pass

2 Answers

Arun
25750 Points
6 years ago
 

We have (1 + x)n = C0 + C1x + C2x2 +...+ Cnxn.                        ... (1)

Multiplying (1) with x, we get

x(1+x)n = C0x + C1x2 + C2x3 +...+ Cnxn+1.                                ... (2)

Differentiating (2) w.r.t. x, we have

(1 + x)n + n(1 + x)n –1 x = C0x + 2C1x2 +...+ (n+1)Cnxn                ... (3)

Putting x = 1 in (3), we get

2n + n.2n –1 = C0 + 2C1 + 3C2 +...+ (n+1)Cn

=> C0 + 2C1 + 3C2 +...+ (n+1)Cn = 2n–1 (n+2). 

hope it helps

Ankit Antony
13 Points
4 years ago
The general term in the above question is 
(r+1)\binom{n}{r}
Therefore we can generalise the whole euation as
\sum_{0}^{n}(r+1)\binom{n}{r}
\sum_{0}^{n}(r)\binom{n}{r}+\sum_{0}^{n}\binom{n}{r}
We know that 
\sum_{0}^{n}\binom{n}{r}=\frac{n}{r}\binom{n-1}{r-1}
\sum_{0}^{n}(r)\binom{n}{r}=\sum_{0}^{n}(r)\frac{n}{r}\binom{n-1}{r-1}
n\sum_{0}^{n}\binom{n-1}{r-1}+\sum_{0}^{n}\binom{n}{r}
Consider,
(1+x)^{n}=\binom{n}{o}+\binom{n}{1}x+\binom{n}{2}x^{2}+.................+\binom{n}{n}x^{n}
Put, x=1
(2)^{n}=\binom{n}{o}+\binom{n}{1}+\binom{n}{2}+..................+\binom{n}{n}
Similarly,
(2)^{n-1}=\binom{n-1}{o}+\binom{n-1}{1}+\binom{n-1}{2}+..................+\binom{n-1}{n-1}
Coming back to our problem
n\sum_{0}^{n}\binom{n-1}{r-1}+\sum_{0}^{n}\binom{n}{r}
Using the given results we obtain ,
n.2^{n-1}+2^{n}
n.2^{n-1}+2^{n-1}.2
2^{n-1}(n+2)
 

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