Simplificar

Question

Simplificando esta expresión $$1\cdot\binom{n}{0}+ 2\cdot\binom{n}{1}+3\cdot\binom{n}{2}+ \cdots+(n+1)\cdot\binom{n}{n}= ?$ $ % $ del $$\text{Hint: } \binom{n}{k}= \frac{n}{k}\cdot\binom{n-1}{k-1} $

Answer 1

Usando la pista, el sumando puede transformarse como $$(k+1){n\choose k}=k{n\choose k}+{n\choose k}=n{n-1\choose k-1}+{n\choose k}.$ $ así obtenemos %#% $ #%

Answer 2

Doble a su expresión y reagrupar usando $\binom{i}{n} = \binom{n-i}{n}$ $$ 1\cdot\binom {n} {0} + 2\cdot\binom {n} {1} + 3\cdot\binom {n} {2} + \cdots+(n+1)\cdot\binom {n} {n} = \\ = \frac 1 2 \left ((1 + (n + 1)) \cdot\binom {n} {0} + (2 + n) \cdot\binom {n} {1} + (3 + (n-1)) \cdot\binom {n} {2} + \cdots+((n+1)+1)\cdot\binom{n}{n}\right) = \frac 1 2 (n + 2) \left (\binom {n} {0} +... + \binom{n}{n}\right) = (n + 2) \cdot 2 ^ {n-1} $$

Answer 3

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