Prove-by-mathematical-induction-that-tbe-following-formula-is-correct-for-all-positive-integers-n-2-2-3-2-4-2-n-1-2-n-2-3-

Question Number 5722 by Rasheed Soomro last updated on 25/May/16

Prove by mathematical induction

that tbe following formula is correct

for all positive integers n :

(\begin{matrix} 2 \\ 2 \end{matrix}) + (\begin{matrix} 3 \\ 2 \end{matrix}) + (\begin{matrix} 4 \\ 2 \end{matrix}) + \dots + (\begin{matrix} n + 1 \\ 2 \end{matrix}) = (\begin{matrix} n + 2 \\ 3 \end{matrix})

Commented by Yozzii last updated on 25/May/16

L e t p (n) : \underset{i = 1}{\sum^{n}} (\begin{matrix} i + 1 \\ 2 \end{matrix}) = (\begin{matrix} n + 2 \\ 3 \end{matrix}) (n \in N)

F o r n = 1

l h s = (\begin{matrix} 1 + 1 \\ 2 \end{matrix}) = (\begin{matrix} 2 \\ 2 \end{matrix}) = 1

r h s = (\begin{matrix} 1 + 2 \\ 3 \end{matrix}) = (\begin{matrix} 3 \\ 3 \end{matrix}) = 1

l h s = r h s \Rightarrow p (n) t r u e f o r n = 1 .

A s s u m e p (n) i s t r u e f o r n = k

\Rightarrow \underset{i = 1}{\sum^{k}} (\begin{matrix} i + 1 \\ 2 \end{matrix}) = (\begin{matrix} k + 2 \\ 3 \end{matrix})

F o r n = k + 1

\Rightarrow \underset{i = 1}{\sum^{k + 1}} (\begin{matrix} i + 1 \\ 2 \end{matrix}) = \underset{i = 1}{\sum^{k}} (\begin{matrix} i + 1 \\ 2 \end{matrix}) + (\begin{matrix} k + 2 \\ 2 \end{matrix})

= (\begin{matrix} k + 2 \\ 3 \end{matrix}) + (\begin{matrix} k + 2 \\ 2 \end{matrix})

= (k + 2)! (\frac{1}{(k - 1)! 3!} + \frac{1}{k (k - 1)! 2!})

= \frac{(k + 2)!}{(k - 1)! 2!} (\frac{1}{3} + \frac{1}{k})

= \frac{(k + 2)! (k + 3)}{k! 3!}

= \frac{(k + 3)!}{(k + 3 - 3)! 3!}

\underset{i = 1}{\sum^{k + 1}} (\begin{matrix} i + 1 \\ 2 \end{matrix}) = (\begin{matrix} k + 3 \\ 3 \end{matrix})

∴ p (k) \Rightarrow p (k + 1) .

∴ p (n) i s t r u e \forall n \in N b y P . M . I .

Commented by Rasheed Soomro last updated on 25/May/16

T h^{a} n k S!