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Grade 12th passAlgebra

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

Vikash Kumar Singh

8 Years agoGrade 12th pass

2 Answers

Arun

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

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

hope it helps

Ankit Antony

7 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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