<s>[INST] Simplify $\sum_{i=0}^n (i+1)\binom ni$ Simplifying this expression$$1\cdot\binom{n}{0}+ 2\cdot\binom{n}{1}+3\cdot\binom{n}{2}+ \cdots+(n+1)\cdot\binom{n}{n}= ?$$
$$\text{Hint: } \binom{n}{k}= \frac{n}{k}\cdot\binom{n-1}{k-1} $$
 [/INST] 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 $$\sum_{k=0}^n(k+1) \dbinom{n}k x^{k} = (n+2) 2^{n-1}$$

</s>