====== 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.

=====Question 10=====
Establish the formulas below by mathematical induction,
$\left(\begin{array}{1}5 \\5 \end{array}\right)+\left(\begin{array}{l}6 \\ 5\end{array}\right)+\left(\begin{array}{l}7 \\ 5\end{array}\right)+\ldots+\left(\begin{array}{c}n+4 \\ 5\end{array}\right)=\left(\begin{array}{c}n+5 \\ 6\end{array}\right)$
====Solution====
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 $n \in \mathbf{N}$.


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