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## 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 $n$ .