# Question 10 Exercise 7.1

Solutions of Question 10 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.

Establish the formulas below by mathematical induction, $\left(\begin{array}{1}5 \\5 \end{array}\right)+\left(\begin{array}{l}6 \\ 5\end{array}\right)+\left(\begin{array}{l}7 \\ 5\end{array}\right)+\ldots+\left(\begin{array}{c}n+4 \\ 5\end{array}\right)=\left(\begin{array}{c}n+5 \\ 6\end{array}\right)$

1. For $n=1$ then \begin{align}\left(\begin{array}{l}5 \\ 5\end{array}\right)&=\dfrac{5 !}{(5-5) ! 5 !}=1\\ \left(\begin{array}{c}1+5 \\ 6\end{array}\right)&=\dfrac{6 !}{(6-6) ! 6 !}=1\end{align} Thus it is true for $n=1$. 2. Let it be true for $n=k$ then \begin{align}\left(\begin{array}{l}5 \\ 5\end{array}\right)+\left(\begin{array}{l}6 \\ 5\end{array}\right)+\left(\begin{array}{l}7 \\ 5\end{array}\right)+\ldots+\left(\begin{array}{c}k+4 \\ 5\end{array}\right)&=\left(\begin{array}{c}k+5 \\ 6\end{array}\right) . ...(i)\end{align} 3. For $n=k+1$ then $(k+1)^{t h}$ term of the series on the left is $a_{k+1}=\left(\begin{array}{c}k+5 \\ 5\end{array}\right)$.

Adding this $a_{k+1}$ term to both sides of the induction hypothesis (i) \begin{align}\left(\begin{array}{l} 5 \\ 5\end{array}\right)+\left(\begin{array}{l} 6 \\ 5\end{array}\right)+\left(\begin{array}{l} 7 \\ 5\end{array}\right)+\ldots+\left(\begin{array}{c} k+4 \\ 5\end{array}\right)+\left(\begin{array}{c} k+5 \\ 5\end{array}\right) \\ & =\left(\begin{array}{c} k+5 \\ 6\end{array}\right)+\left(\begin{array}{c}k+5 \\5 \end{array}\right) \\ & =\dfrac{(k+5) !}{(k+5-6) ! 6 !}+\dfrac{(k+5) !}{(k+5-5) ! 5 !} \\ & =\dfrac{(k+5) !}{(k-1) ! 6.5 !}+\dfrac{(k+5) !}{k ! 5 !}\\ & =\dfrac{(k+5) !}{(k-1) ! 6.5 !}+\dfrac{(k+5) !}{k(k-1) ! 5 !} \\ & =\dfrac{(k+5) !}{(k-1) ! 5 !}\left[\dfrac{1}{6}+\dfrac{1}{k}\right] \\ & =\dfrac{(k+5) !}{(k-1) ! 5 !}\left[\dfrac{k+6}{6 k}\right] \\ & =\dfrac{(k-6)(k+5) !}{k(k-1) ! 6.5 !} \\ & =\dfrac{(k+6) !}{k ! 6 !} \\ \Rightarrow\left(\begin{array}{c} 5 \\ 5 \end{array}\right)+\left(\begin{array}{c} 6 \\ 5 \end{array}\right)+\left(\begin{array}{l} 7 \\ 5 \end{array}\right)+\ldots+\left(\begin{array}{c} k-4 \\ 5 \end{array}\right) \\ & +\left(\begin{array}{c} k+5 \\ 5 \end{array}\right)&=\left(\begin{array}{c} k+1-5 \\ 6 \end{array}\right)\end{align} Which is the just the form taken by given propusition when $n$ is replaced by $k+1$.

hence it is true for $n=k+1$. Thus by mathematical induction it is true for all $n \in \mathbf{N}$.