====== Question 11 Exercise 7.1 ======
Solutions of Question 11 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 11=====
\begin{align}
& \left(\begin{array}{l}
2 \\
2
\end{array}\right)+\left(\begin{array}{l}
3 \\
2
\end{array}\right)+\left(\begin{array}{l}
4 \\
2
\end{array}\right)+\ldots+\left(\begin{array}{l}
n \\
2
\end{array}\right)=\left(\begin{array}{c}
n+1 \\
3
\end{array}\right), \quad n \geq 2
\end{align}
====Solution====
1. For $n=2$ then
\begin{align}
\left(\begin{array}{l}
2 \\
2
\end{array}\right)&=\dfrac{2 !}{(2-2) ! 2 !}=1 \\
\left(\begin{array}{c}
2+1 \\
2
\end{array}\right)&=\dfrac{3 !}{(3-3) ! 3 !}=1\end{align}
Thus it is true for $n=1$.

2. Let it be true for $n=k$ then
$$\left(\begin{array}{l}
2 \\
2
\end{array}\right)+\left(\begin{array}{l}
3 \\
2
\end{array}\right)+\left(\begin{array}{l}
4 \\
2
\end{array}\right)+\ldots+\left(\begin{array}{l}
k \\
2
\end{array}\right)=\left(\begin{array}{c}
k-1 \\
3
\end{array}\right)$$
3. For $n=k+1$ then $(k+i)^{u / i}$ term of the series on the left is $a_{k-1}=\left(\begin{array}{c}k-1 \\ 2\end{array}\right)$.

Adding this $a_{k+1}$ to both sides of the induction hypothesis, we have
\begin{align}(\begin{array}{l}
2 \\
2
\end{array})+(\begin{array}{l}
3 \\
2
\end{array})+(\begin{array}{l}
4 \\
2
\end{array})+\ldots+(\begin{array}{l}
k \\
2
\end{array})+(\begin{array}{c}
k+1 \\
2
\end{array})
& =\left(\begin{array}{c}
k+1 \\
3
\end{array}\right)+\left(\begin{array}{c}
k+1 \\
2
\end{array}\right) \\
& =\dfrac{(k+1) !}{(k+1-3) ! 3 !}+\dfrac{(k+1) !}{(k+1-2) ! 2 !} \\
& =\dfrac{(k+1) !}{(k-2) ! 3 !}+\dfrac{(k+1) !}{(k-1) ! 2 !} \\
& =\dfrac{(k+1) !}{(k-2) ! 3 \cdot 2 !}+\dfrac{(k+1) !}{(k-1)(k-2) ! 2 !} \\
& =\dfrac{(k+1) !}{(k 2) ! 2 !}\left[\dfrac{1}{3}+\frac{1}{k-1}\right] \\
& =\dfrac{(k+1) !}{\left(k_2\right) ! 2 !}\left[\dfrac{k-1+3}{3(k-1)}\right] \\
& =\dfrac{(k+2)(k+1) !}{(k-1)(k-2) ! 3 \cdot 2 !} \\
& =\dfrac{(k+2) !}{(k-1) ! 3 !} \\
\Rightarrow(\begin{array}{l}
2 \\
2
\end{array})+(\begin{array}{l}
3 \\
2
\end{array})+(\begin{array}{l}
4 \\
2
\end{array})+\ldots+(\begin{array}{l}
k \\
2
\end{array})+(\begin{array}{c}
k+1 \\
2
\end{array})
& =(\begin{array}{c}
k+2 \\
3
\end{array})\\&=(\begin{array}{c}
k+1+1 \\
3
\end{array})\end{align}
Which is the form of the proposition when $n$ is replaced by $k+1$,

hence it is true for $n=k+1$. 

Thus by mathematical induction it is true for all $n \geq 2$.





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