Question

Prove that $\mathbf{\sum }_{\mathbf{j}\mathbf{=}\mathbf{0}}^{\mathbf{n}}\mathbf{ }{\left(\frac{-1}{2}\right)}^{\mathbf{j}}\mathbf{=}\frac{{\mathbf{2}}^{\mathbf{n}\mathbf{+}\mathbf{1}}\mathbf{+}\mathbf{(}\mathbf{-}\mathbf{1}{\mathbf{)}}^{\mathbf{n}}}{\mathbf{3}\mathbf{-}{\mathbf{2}}^{\mathbf{n}}}$ whenever n is a nonnegative integer.

Short answer

It is proved that $\sum _{j=0}^n {\left(\frac{-1}{2}\right)}^j=\frac{2^{n+1}+(-1{)}^n}{3\cdot 2^n}$.

Step-by-step solution

Mathematical Induction

The principle of mathematical induction is to prove that P(n) is true for all positive integer n in two steps.

1. Basic step : To verify that P(1) is true.

2. Inductive step : To prove the conditional statement if P(k) is true then P(k+1) is true.

Proof by induction

Let P(n) be the statement $\sum _{j=0}^n {\left(\frac{-1}{2}\right)}^j=\frac{2^{n+1}+(-1{)}^n}{3\cdot 2^n}$. Let us prove by induction on n.

Base Case:

For n = 0 , the value of LHS is 1 and RHS is 1. Since, both are equal the statement is true for P(1) .

Induction Case:

Assume that the statement is true for P(k) then prove for P(k+1) .

Let P(k) be true then $\sum _{j=0}^k {\left(\frac{-1}{2}\right)}^j=\frac{2^{k+1}+(-1{)}^k}{3\cdot 2^k}$. Let us prove the statement is true for P(k+1) as follows:

$\begin{array}{r}\sum _{j=0}^{k+1} {\left(\frac{-1}{2}\right)}^j=\sum _{j=0}^k {\left(\frac{-1}{2}\right)}^j+{\left(\frac{-1}{2}\right)}^{k+1}\\ =\frac{2^{k+1}+(-1{)}^k}{3\cdot 2^k}+\frac{(-1{)}^{k+1}}{2^{k+1}}\\ =\frac{2\left(2^{k+1}+(-1{)}^k\right)}{3\cdot 2^{k+1}}+\frac{3\cdot (-1{)}^{k+1}}{3\cdot 2^{k+1}}\\ =\frac{2^{k+2}+2(-1{)}^k+3\cdot (-1{)}^{k+1}}{3\cdot 2^{k+1}}\end{array}$

Further, simplify the values as follows:

$\begin{array}{r}\sum _{j=0}^{k+1} {\left(\frac{-1}{2}\right)}^j=\frac{2^{k+2}+2(-1)(-1{)}^{k+1}+3\cdot (-1{)}^{k+1}}{3\cdot 2^{k+1}}\\ =\frac{2^{k+2}+(-1{)}^{k+1}(-2+3)}{3\cdot 2^{k+1}}\\ =\frac{2^{k+2}+(-1{)}^{k+1}(1)}{3\cdot 2^{k+1}}\\ =\frac{2^{(k+1)+1}+(-1{)}^{k+1}}{3\cdot 2^{k+1}}\end{array}$

Thus, P(k+1) is also true.

Therefore, the statement P(n) is true for every positive integer n.