Question

(a) Find the formula for $\frac{\mathbf{1}}{\mathbf{2}}\mathbf{+}\frac{\mathbf{1}}{\mathbf{4}}\mathbf{+}\frac{\mathbf{1}}{\mathbf{8}}\mathbf{+}\mathbf{\cdots }\mathbf{+}\frac{\mathbf{1}}{{\mathbf{2}}^{\mathbf{n}}}$ by examining the values of this expression for small values of n.

(b) Prove the formula you conjectured in part (a).

Short answer

(a)The general formula for the given series is $\frac{2^n-1}{2^n}$.

(b) The statement P(n) is true for every positive integer n.

Step-by-step solution

Formula for the given series

Consider the given series as$f\left(n\right)=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\cdots +\frac{1}{2^n}$ …… (1)

For n = 1 ,$\frac{1}{2^1}=\frac{1}{2}$ .

For n = 2 ,

$\begin{array}{r}\frac{1}{2}+\frac{1}{2^2}=\frac{2+1}{2^2}\\ =\frac{3}{2^2}\\ =\frac{2^2-1}{2^2}\end{array}$

For n = 3 ,

$\begin{array}{r}\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}=\frac{2^2+2+1}{2^3}\\ =\frac{7}{2^3}\\ =\frac{2^3-1}{2^3}\end{array}$

Therefore, the general formula for the given series is $\frac{2^n-1}{2^n}$.

Proof for the Formula

Let P(n) be the statement $\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\cdots +\frac{1}{2^n}=\frac{2^n-1}{2^n}$.

Let us prove by induction on n.

Base Case:

For n = 1 , the value of LHS is $\frac{1}{2}$ and RHS is $\frac{1}{2}$. Since, both are equal the statement is true for

Induction Case:

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

Let P(k) be true as $\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\cdots +\frac{1}{2^k}=\frac{2^k-1}{2^k}$. Let us prove the statement is true for as follows:

$\begin{array}{r}\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\cdots +\frac{1}{2^k}+\frac{1}{2^{k+1}}=\frac{2^k-1}{2^k}+\frac{1}{2^{k+1}}\\ =\frac{2\left(2^k-1\right)}{2^{k+1}}+\frac{1}{2^{k+1}}-\\ =\frac{2^{k+1}-2+1}{2^{k+1}}\\ =\frac{2^{k+1}-1}{2^{k+1}}\end{array}$

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

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