Question

Prove part (b) of theorem 4.4 in the case when n is even: if n is a positive even integer, then$\sum _{k=1}^{n}k=\frac{n(n+1)}{2}$

Short answer

We proved$\sum _{k=1}^{n}k=\frac{n(n+1)}{2}$

Step-by-step solution

Given information

We are given a relation we have to prove it

$\sum _{k=1}^{n}k=\frac{n(n+1)}{2}$, Where n is even

Prove using mathematical induction

First check whether the above relation is true for n=2

We have,

LHS=3

And

$$\begin{gathered} RHS=\frac{n(n+1)}{2} \\ RHS=\frac{2(2+1)}{2} \\ RHS=3 \end{gathered}$$

Hence LHS=RHS

Now consider that the above statement is true for n where n is an even number

We have

$\sum _{k=1}^{n}k=\frac{n(n+1)}{2}$

Now we prove that the statement is true for n+2

That is

$\sum _{k=1}^{n+2}k=\frac{(n+2)(n+3)}{2}$

Consider LHS

role="math" localid="1648555801521" $$\begin{gathered} \sum _{k=1}^{n+2}k \\ =\sum _{k=1}^{n}k+n+1+n+2 \\ =\frac{n(n+1)}{2}+2n+3 \\ =\frac{n(n+1)+4n+6}{2} \\ =\frac{n^2+n+5n+6}{2} \\ =\frac{(n+2)(n+3)}{2} \end{gathered}$$

Hence proved