Solution to a) i) Give an inductive formula for the sum of the first n odd numbers: … - Sikademy
Author Image

Archangel Macsika

a) i) Give an inductive formula for the sum of the first n odd numbers: 1 + 3 + 5 + ... + 2n -1 Show your induction process. ii) Use the proof by mathematical induction to prove the correctness of your inductive formula in i) above.

The Answer to the Question
is below this banner.

Can't find a solution anywhere?

NEED A FAST ANSWER TO ANY QUESTION OR ASSIGNMENT?

Get the Answers Now!

You will get a detailed answer to your question or assignment in the shortest time possible.

Here's the Solution to this Question

a)

i)


a_1=1, a_2=3, a_n=2n-1=1+2(n-1)

We have the arithmetic progression with the first term a_1=1 and the common difference d=2.


S_n=\dfrac{a_1+a_n}{2}\cdot n=\dfrac{1+2n-1}{2}\cdot n=n^2

ii) Let P(n) be the proposition that the sum of the first n odd positive integers, 1+3+5+...+2n-1 is n^2.

Basis Step

P(1) is true, because 1=1^2.


Inductive Step

For the inductive hypothesis we assume that P(k) holds for an arbitrary positive integer k. That is, we assume that


1+3+5+...+2k-1=k^2

Under this assumption, it must be shown that P(k + 1) is true, namely, that


1+3+5+...+2k-1+2(k+1)-1=(k+1)^2

is also true.

When we add k + 1 to both sides of the equation in P(k), we obtain


1+3+5+...+2k-1+2(k+1)-1=k^2+2(k+1)-1

=k^2+2k+1

=(k+1)^2

This last equation shows that P(k + 1) is true under the assumption that P(k) is true. This completes the inductive step.

We have completed the basis step and the inductive step, so by mathematical induction we know that P(n) is true for all positive integers n. That is, we have proven that the sum of the first n odd numbers 1+3+5+...+2n-1=n^2 for all positive integres n .


Related Answers

Was this answer helpful?

Join our Community to stay in the know

Get updates for similar and other helpful Answers

Question ID: mtid-5-stid-8-sqid-185-qpid-73