# Question 7 Exercise 7.1

Solutions of Question 7 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 7

Establish the formulas below by mathematical induction, $1.2+2.3+3.4+\ldots+n(n+1)=\dfrac{n(n+1)(n+2)}{3}$

### Solution

1. For $n=1$ then $$1.2=2=\dfrac{1(1+1)(1+2)}{3}=2 $$ Thus it is true for $n=1$.

2. Let it be true for $n=k$, then \begin{align}1.2+2.3+3.4+\ldots+k(k+1)& =\dfrac{k(k+1)(k+2)}{3}....(i)\end{align} 3. Considering for $n=k+1$, then $(k-1)^{t h}$ term of the series on left is $a_{k+1}=(k+1)(k+ 2)$.

Adding this $(k+1)^{\text {th }}$ term to both sides of the induction hypothesis (i), we have \begin{align}1.2+2.3+3.4+\ldots+k(k+1)+(k+1)(k+2) & =\dfrac{k(k+1)(k-2)}{3}+(k+1)(k+2) \\ & =(k-1)(k+2)[\dfrac{k}{3}+1] \\ & =(k-1)(k+2) \dfrac{k+3}{3} \\ \Rightarrow 1.2+2.3+3.4+\ldots+k(k+1)+(k+1)(k+2)& =\dfrac{(k+1)(k+1+1)(k+1+2)}{3}\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 $n \in \mathbf{N}$.

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