====== 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.

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





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