# Question 5 Exercise 7.1

Solutions of Question 5 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 5

Establish the formulas below by mathematical induction, $1^3+2^3+3^3+\ldots+n^3=\left[\dfrac{n(n+1)}{2}\right]^2$.

### Solution

1. For $n=1$, then $1^3=1=\left[\dfrac{1(1+1)}{2}\right]^2=1$.

Thus it is true for $n=1$.

2. Let it be true for $n=k_1$, then \begin{align}1^3+2^3+3^3+\ldots+k^3& =[\dfrac{k(k+1)}{2}]^2....(i)\end{aligned} 3. Now $n=k+1$ the $(k+1)$ term of the given series on left is $a_{k+1}=(k+1)^3$.

Adding this $(k+1)^{t h}$ term to both sides of the induction hypothesis (i), we have \begin{align}1^3+2^3+3^3+\ldots+k^3+(k+1)^3 & =[\dfrac{k(k+1)}{2}]^2+(k+1)^3 \\ & =(k+1)^2[\dfrac{k^2}{4}+(k+1)] \\ & =(k+1)^2[\dfrac{k^2+4(k+1)}{4}] \\ & =\dfrac{(k-1)^2(k^2+4 k+4)}{4}-\dfrac{(k+1)^2(k+2)^2}{2^2} \\ \Rightarrow 1^3+2^3-3^3+\ldots+k^3+(k+1)^3 & =\left[\dfrac{(k+1)(k+1+1)}{2}\right]^2\end{align} Which is the form taken ny 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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