Question 4 Exercise 7.1
Solutions of Question 4 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 4
Establish the formulas below by mathematical induction 3+7+11+⋯+(4n−1)=n(2n+1)
Solution
1. For n=1 then 3=1(2+1)=3 thus the above statement or proposition is true for n=1.
2. Let it be true for n=k, we have 3+7+11+⋯+(4k−1)=k(2k+1)....(i) 3. Now considering for n=k+1, the (k+1) term of the series is ak+1=4(k+1)−1.
Adding this (k+1)th term to both sides of the induction hypothesis (i), we have 3+7+11+⋯+(4k−1)+[4(k+1)−1]=k(2k+1)+4(k+1)−1=2k2+k+4k+4−1=2k2+5k+3=2k2+2k+3k+3=2k(k+1)+3(k+1)=(k+1)(2k+3)=(k+1)[2k+2+1]=(k+1)[2(k+1)+1]3+7+11+⋯+(4k−1)+[4(k+1)−1]=(k+1)[2(k+1)+1] Which is the form of given statement 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∈N.
Go To