Answer to Question #299223 in Combinatorics | Number Theory for Maneo

Question #299223

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.


1
Expert's answer
2022-02-18T15:47:17-0500

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" .



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