Chapter 4: Q. 61 (page 327)

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

Expert verified

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

$R H S = \frac{n (n + 1)}{2} R H S = \frac{2 (2 + 1)}{2} R H S = 3$

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" $\sum_{k = 1}^{n + 2} k = \sum_{k = 1}^{n} k + n + 1 + n + 2 = \frac{n (n + 1)}{2} + 2 n + 3 = \frac{n (n + 1) + 4 n + 6}{2} = \frac{n^{2} + n + 5 n + 6}{2} = \frac{(n + 2) (n + 3)}{2}$

Hence proved

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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

Step by step solution

Given information

Prove using mathematical induction

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Prove part (b) of theorem 4.4 in the case when n is even: if n is a positive even integer, then∑k=1nk=n(n+1)2

Short Answer

Step by step solution

Given information

Prove using mathematical induction

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Geometry

Decision Maths

Theoretical and Mathematical Physics

Pure Maths

Applied Mathematics

Calculus

Study anywhere. Anytime. Across all devices.

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