# Question 6 Exercise 7.1

Solutions of Question 6 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 6

Establish the formulas below by mathematical induction, $1(1 !)+2(2 !)+3(3 !)+\ldots+n(n !)= -(n+1) !-1$

### Solution

1. For $n=1$, then $$1(1 !)=1=(1+1) !-1=2 !-1=1 $$ Thus it is truc for $n=1$.

2. Let it be true for $n=k$, then we have \begin{align}1(1 !)+2(2 !)+3(3 !)+\ldots+k(k !)& =(k+1) !-1 \ldots . .(i)\end{align} 3. For $n=k+1$ the $(k+1)^{t h}$ term of the series on the left is $a_{k+1}=(k+1)[(k+1) !]$.

Actding this $a_{k-1}$ term to both sides of the induction hypothesis(i), we have \begin{align}1(1 !)+2(2 !)+3(3 !)+\ldots+k(k !)+ (k+1) !(k+1) ! & =(k+1) !-1-k+1,[k+1) ! \\ & =(k+1) !+(k+1)[(k+1) ![-1 \\ & =(k+1) ![1+k+1]-1 \\ & =(k+1) !(k+2)-1 \\ & =(k+2)(k+1) !-1 \\ & =(k+2) !-1 \quad \text { as }(k+2)(k+1) !=(k+2) ! \\ \rightarrow 1(1 !)+2(2 !)+3(3 !)+\ldots+k(k !)+(k+1)[(k+1) !] & =(k+2) !-1=(k+1+1) !-1 \end{align} Which is the form 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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