# Question 9 Exercise 7.1

Solutions of Question 9 of Exercise 7.1 of Unit 07: Permutation, Combination and Probablity. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.

Establish the formulas below by mathematical induction, $\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+\ldots+\dfrac{1}{3^n}=\dfrac{1}{2}[1-\dfrac{1}{3^n}]$

1. For $n=1$ then $$\dfrac{1}{3}-\dfrac{1}{2}[1-\dfrac{1}{3}]-\dfrac{1}{2} \dfrac{2}{3}=\dfrac{1}{3}$$ Thus it is true for $n=1$.

2. Let it be true for $n=k$ then $$\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+\ldots+\dfrac{1}{3^k}-\dfrac{1}{2}[1-\dfrac{1}{3^k}]$$ 3. For $n=k+1$, the $(k+1)^{t h}$ term of the series on left is $a_{k+1}=\frac{1}{3^{k+1}}$.

Adding this $a_{k+1}$ term to both sides of the induction hypothesis \begin{align}\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+\ldots+\dfrac{1}{3^k}+\dfrac{1}{3^{k+1}} \\ & =\dfrac{1}{2}[1-\dfrac{1}{3^k}]+\dfrac{1}{3^{k+1}}\\ & =\dfrac{1}{2}-\dfrac{1}{2 \cdot 3^k}+\dfrac{1}{3 \cdot 3^k} \\ & =\dfrac{1}{2}+\dfrac{1}{3^k}(\dfrac{1}{3}-\dfrac{1}{2}) \\ & =\dfrac{1}{2}+\dfrac{1}{3^k} \dfrac{2-3}{2 \cdot 3} \\ & =\dfrac{1}{2} \dfrac{1}{2 \cdot 3 \cdot 3^k} \\ \Rightarrow \dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+\ldots+\dfrac{1}{3^k}+\dfrac{1}{3^{k+1}} & =\dfrac{1}{2}\left[1-\dfrac{1}{3^{k-1}}\right]\end{align} Which is the form taken by proposition when $n$ is replaced by $k+1$. hence it is true for $n=k+1$.

Thus by mathematical induction it is true for all $n \leq \mathbf{N}$.