====== 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 $3+7+11+\cdots+(4 n-1)=n(2 n+1)$
====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
\begin{align}3+7+11+\cdots+(4 k-1) 
& =k(2 k+1)....(i) \end{align}
3. Now considering for $n=k+1$, the $(k+1)$ term of the series is $a_{k+1}=4(k+1)-1$.

Adding this $(k+1)^{t h}$ term to both sides of the induction hypothesis (i), we have
\begin{align}
3+7+11+\cdots+(4 k-1)+[4(k+1)-1] & =k(2 k+1)+4(k+1)-1 \\
& =2 k^2+k+4 k+4-1 \\
& =2 k^2+5 k+3 \\
& =2 k^2+2 k+3 k+3 \\
& =2 k(k+1)+3(k+1) \\
& =(k+1)(2 k+3) \\
& =(k+1)[2 k+2+1] \\
& =(k+1)[2(k+1)+1] \\
3+7+11+\cdots+(4 k-1)+[4(k+ 1)-1]&=(k+1)[2(k+1)+1]\end{align}
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 $n \in \mathbf{N}$.






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