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

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question markProve that C0 +2C1 + 3C2 +....... + (n+1)Cn = 2^(n-1) × (n+2)
Vikash Kumar Singh , 8 Years ago
Grade 12th pass anser 2 Answers
Arun

Last Activity: 8 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

Last Activity: 6 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 problemn\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}.22^{n-1}(n+2) star
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