A closed form for $\sum_{i=1}^{n} i\cdot2^i$

Question

Let $f(n) = \sum_{i=1}^{n} i\cdot(2^i)$ How to write $f(n)$ only in terms of $n$.

I am unable to find. Thanks in advance.

Answer 1

Hint

Note that $$ \frac{x(1-x^n)}{1-x}=\sum_{i=1}^{n}x^i. $$ Hence $$ x\frac{d}{dx}\left(\frac{x(1-x^n)}{1-x}\right)=\sum_{i=1}^{n}ix^i $$ You should be able to take it from here.

Answer 2

An elementary approach based upon the geometric series summation formula is \begin{align*} \color{blue}{\sum_{i=1}^ni2^i}&=\sum_{i=1}^n\left(\sum_{j=1}^i1\right)2^i=\sum_{i=1}^n\sum_{j=1}^i2^i\tag{1}\\ &=\sum_{1\leq j\leq i\leq n}2^i\\ &=\sum_{j=1}^n\sum_{i=j}^n2^i=\sum_{j=1}^n\sum_{i=0}^{n-j}2^{i+j}\tag{2}\\ &=\sum_{j=1}^n2^j\sum_{i=0}^{n-j}2^{i}=\sum_{j=1}^n2^j\cdot\frac{2^{n-j+1}-1}{2-1}\tag{3}\\ &=\sum_{j=1}^n2^{n+1}-\sum_{j=1}^n2^j\tag{4}\\ &=n2^{n+1}-\left(\frac{2^{n+1}-1}{2-1}-1\right)\tag{5}\\ &\color{blue}{=2^{n+1}(n-1)+2}\tag{6} \end{align*}

Comment:

• In (1) we represent the factor $i$ as sum.

• In (2) we exchange the order of summation and shift the inner index $i$ to start from $i=0$.

• In (3) we apply the geometric series summation formula to the inner sum.

• In (4) we simplify and split the sum.

• In (5) we calculate both sums by again applying the geometric series summation formula to the right-hand sum.

• In (6) we do some final simplifications.