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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