Question 3 Exercise 7.1

Solutions of Question 3 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.

Establish the formulas below by mathematical induction $3+6+9+\ldots+3 n=\dfrac{3 n(n+1)}{2}$

1. For $n=1$ then $3=\dfrac{3.1(1+1)}{2}=3$,

thus the above statement or proposition is true for $n=1$.

2. Let it be true for $n=k$, we have $$3+6+9+\ldots+3 k=\dfrac{3 k(k+1)}{2}....(i)$$ 3. Now considering for $n=k+1$, the $(k+1)$ term of the series is $a_{k+1}=3(k+1)$.

Adding this $(k+1)^{t h}$ term to both sides of the induction hypothesis (i), we have \begin{align}3+6+9+\ldots+3 k+3(k+1) & =\dfrac{3 k(k+1)}{2}+3(k+1) \\ & =3(k+1)[\dfrac{k}{2}+1] \\ & =3(k+1)[\dfrac{k+2}{2}] \\ \Rightarrow 3+6+9+\ldots+3 k+3(k+1) & =\dfrac{3(k+1)(k+1-1)}{2}\end{align} Which is the form of given statement 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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