Question

Use mathematical induction to show that

$\frac{\mathbf{2}}{\mathbf{3}}\mathbf{+}\frac{\mathbf{2}}{\mathbf{9}}\mathbf{+}\frac{\mathbf{2}}{\mathbf{27}}\mathbf{+}\mathbf{\dots }\mathbf{+}\frac{\mathbf{2}}{{\mathbf{3}}^{\mathbf{n}}}\mathbf{=}\mathbf{1}\mathbf{-}\frac{\mathbf{1}}{{\mathbf{3}}^{\mathbf{n}}}$whenever nis a positive integer.

Short answer

It is prove that$\frac{2}{3}+\frac{2}{9}+\frac{2}{27}+\dots +\frac{2}{3^n}=1-\frac{1}{3^n}$ for nis a positive integer.

Step-by-step solution

Principle of Mathematical Induction 

Consider the propositional function P (n). Consider two actions to prove that P (n) evaluates to accurate for all set of positive integers n.

Consider the first basic step is to confirm that P(1) true.

Consider the inductive step is to demonstrate that for any positive integer k the conditional statement$\mathbf{P}\mathbf{\left(}\mathbf{k}\mathbf{\right)}\mathbf{\to }\mathbf{P}\mathbf{(}\mathbf{k}\mathbf{+}\mathbf{1}\mathbf{)}$is true.

Prove the basis step

Given statement is

$\frac{2}{3}+\frac{2}{9}+\frac{2}{27}+\dots +\frac{2}{3^n}=1-\frac{1}{3^n}$

In the basis step, we need to prove that P (1) is true.

For finding statement P(1) substituting for n in the statement.

Therefore, the statement P (1) is:

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

The statement P (1) is true this is also known as the basis step of the proof.

Prove the Inductive step

In the inductive step, we need to prove that, if P (k) is true, then P (k + 1) is also true.

That is,

$P\left(k\right)\to P(k+1)$ is true for all positive integers k.

In the inductive hypothesis, we assume that P (k) is true for any arbitrary positive integer k

$\frac{2}{3}+\frac{2}{9}+\frac{2}{27}+\dots +\frac{2}{3^k}=1-\frac{1}{3^k}$ .........(1)

Now we must have to show that P (k + 1) is also true

Therefore, replacing k with k + 1 in the statement.

$\frac{2}{3}+\frac{2}{9}+\frac{2}{27}+\dots +\frac{2}{3^{k+1}}=1-\frac{1}{3^{k+1}}$

Now, adding $\frac{2}{3^{k+1}}$in both sides of the equation (i) or inductive hypothesis.

$\begin{array}{r}\frac{2}{3}+\frac{2}{9}+\frac{2}{27}+\dots +\frac{2}{3^{\mathrm{k}}}+\frac{2}{3^{\mathrm{k}+1}}=1-\frac{1}{3^{\mathrm{k}}}+\frac{2}{3^{\mathrm{k}+1}}\\ =1-\left(\frac{1}{3^{\mathrm{k}}}-\frac{2}{3^{\mathrm{k}+1}}\right)\\ =1-\frac{3-2}{3^{\mathrm{k}+1}}\\ =1-\frac{1}{3^{\mathrm{k}+1}}\end{array}$

From the above, we can see that is also true

Hence , P (k + 1)is true under the assumption that P (k)is true. This

completes the inductive step.

Hence it is shown that$\frac{2}{3}+\frac{2}{9}+\frac{2}{27}+\dots +\frac{2}{3^n}=1-\frac{1}{3^{n+1}}$whenever nis positive integer.