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Simplifying this expression$$1\cdot\binom{n}{0}+ 2\cdot\binom{n}{1}+3\cdot\binom{n}{2}+ \cdots+(n+1)\cdot\binom{n}{n}= ?$$ $$\text{Hint: } \binom{n}{k}= \frac{n}{k}\cdot\binom{n-1}{k-1} $$

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We have $$\sum_{k=0}^n\dbinom{n}k x^k = (1+x)^n$$ Hence, we get that $$\sum_{k=0}^n\dbinom{n}k x^{k+1} = x(1+x)^n$$ Differentiating the above, we get $$\sum_{k=0}^n(k+1) \dbinom{n}k x^{k} = \dfrac{d(x(1+x)^n)}{dx}$$ Now set $x=1$ to get the answer.

$$\sum_{k=0}^n(k+1) \dbinom{n}k x^{k} = \dfrac{d(x(1+x)^n)}{dx} = (1+x)^{n-1} \left(nx+x+1\right)$$ Setting $x=1$, we get that $$\sum_{k=0}^n(k+1) \dbinom{n}k x^{k} = (n+2) 2^{n-1}$$

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  • $\begingroup$ Why did you differentiate? $\endgroup$ – soandos Apr 4 '13 at 17:49
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Using the hint, the summand can be transformed as $$(k+1){n\choose k}=k{n\choose k}+{n\choose k}=n{n-1\choose k-1}+{n\choose k}.$$ Thus we obtain $$\sum_{k=0}^n(k+1){n\choose k} = n\sum {n-1\choose k-1}+\sum {n\choose k} = n2^{n-1}+2^n. $$

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Just double your expression and regroup using $\binom{i}{n} = \binom{n-i}{n}$ $$ 1\cdot\binom{n}{0}+ 2\cdot\binom{n}{1}+3\cdot\binom{n}{2}+ \cdots+(n+1)\cdot\binom{n}{n}= \\ =\frac 1 2 \left( (1 + (n+1))\cdot\binom{n}{0}+ (2 + n)\cdot\binom{n}{1}+(3 + (n-1))\cdot\binom{n}{2}+ \cdots+((n+1)+1)\cdot\binom{n}{n}\right)= \frac 1 2 (n + 2)\left(\binom{n}{0} + ... + \binom{n}{n}\right) = (n+2) \cdot 2^{n-1} $$

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\begin{align}\sum_{k=0}^n(k+1) \binom{n}{k}&=\sum_{k=0}^nk \binom{n}{k}+\sum_{k=0}^n \binom{n}{k}=n\sum_{k=1}^n \binom{n-1}{k-1}+2^n\\&=n\sum_{k=0}^{n-1} \binom{n-1}{k}+2^n=n2^{n-1}+2^n \end{align}

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$$\sum_{i=0}^n (i+1)\binom ni=\sum_{i=0}^n i\binom ni+\sum_{i=0}^n \binom ni=\sum_{i=0}^n i\frac ni\binom {n-1}{i-1}+2^n=$$ $$=n\sum_{i=0}^n\binom {n-1}{i-1}+2^n=n\sum_{j=0}^n\binom {n-1}{j}+2^n=n2^{n-1}+2^n$$

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