Question 8 Exercise 7.1
Solutions of Question 8 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 8
Establish the formulas below by mathematical induction, $1+2+2^2+2^3+\ldots+2^n 1=2^n-1$.
Solution
1. For $n=1$, we have $1=2^1-1=1$.
Thus it is true for $n=1$.
2. Let it be true for $n-k>1$ then \begin{align}1+2+2^2+2^3+\ldots+2^{k-1} \\ & =2^k-1 ....(i)\end{align} 3. Considering for $n-k-1$, then $(k+1)^{t h}$ term of the series on the left is $a_{k+1}=2^k$.
Adding this $a_{k+1}$ term to both sides of the induction hypothesis (i), we have \begin{align}1+2+2^2+2^3+\ldots+2^{k-1}-2^k & =2^k-12^k \\ & =2^k+2^k-1=2.2^k-1 \\ \Rightarrow & 1+2+2^2+2^3+\ldots+2^{k-1}+2^k & =2^{k+1}-1\end{align} Which is the form of the 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 positive integers.
Go To