====== Question 2 Exercise 7.1 ======
Solutions of Question 2 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 2=====
Establish the formulas below by mathematical induction, $1+5+9+\ldots+(4 n-3)=n(2 n-1)$
====Solution====
1. For $n=1$ then
$$1=1(2.1-1)=1$$
Hence the above proposition is true for $n=1$.

2. Let it be true for $n=k$, then
\begin{align}1+5+9+\ldots+(4 k-3)\\
& =k(2 k-1)....(i) \\
\end{align}
3. For $n=k+1$, the $k+1$ term of the series, which is:
$$a_{k-1}=4(k+1)-3=4 k+1 $$
Adding this $(k+1)^{t h}$ term to both sides of the induction hypcthesis (i)
\begin{align}1+5+9+\ldots+(4 k-3)+(4 k+1)& =k(2 k-1)+4 k+1 \\
& =2 k^2-k+4 k+1 \\
& =2 k^2+3 k+1 \\
& =2 k^2+2 k+k+1 \\
& =2 k(k+1)+1(k+1) \\
\Rightarrow 1+5+9+\ldots+(4 k-3)+(4 k+1)& =(k+1)[2 k+1] \\
& =(k+1)[2(k+1)-1]\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 it true for $n \in \mathbf{N}$.





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