Proving arithmetic series by induction

Question

How do I prove this statement by the method of induction:

$$ \sum_{r=1}^n [d + (r - 1)d] = \frac{n}{2}[2a + (n - 1)d] $$

I know that $d + (r - 1)d$ stands for $u_n$ in an arithmetic series, and the latter statement represents the sum of the series, but I'm not sure how to prove them by induction.

If I let $n = k$:

$$ \sum_{r=1}^k [d + (r - 1)d] = \frac{k}{2}[2a + (k - 1)d] $$

the next step is to realize that the following term, $k + 1$, is equated to:

$$ \sum_{r=1}^{k+1} [d + (r - 1)d] = \sum_{r=1}^k [d + (r - 1)d] + [d + (k + 1 - 1)d] $$ $$ = \frac{k}{2}[2a + (k - 1)d] + d + dk $$ $$ = \frac{k(2a + dk - d) + 2d + 2dk}{2} $$ $$ = \frac{2ak + dk^2 - dk + 2d + 2dk}{2} $$ $$ = \frac{dk^2 + dk + 2ak + 2d}{2} $$

What's the next step I take? Do I substitute back in $k = n - 1$?

Answer 1

There is an easier way. Notice that: $$\sum_{r = 1}^n a + (r-1)d = \sum_{r = 1}^n a + \sum_{r = 1}^n rd - \sum_{r = 1}^n d = na + d\left (\sum_{r = 1}^n r\right) - nd $$

Now, just prove by induction that $$\sum_{r = 1}^n r = \frac{n(n+1)}{2}$$ which is much easier, and manipulate the previous expression to get what you need.

Answer 2

The $r$th term, $u_r=a+(r-1)d$ with $u_1=a$ be the first term & $d$ be the common difference

If $\displaystyle S_n= \sum_{r=1}^nu_n=\frac n2[2a+(n-1)d]$

$\displaystyle\implies S_{n+1}=\displaystyle \sum_{r=1}^{n+1}u_n=S_n+u_{n+1}=\frac n2[2a+(n-1)d]+a+nd$ $\displaystyle\implies S_{n+1}=(n+1)a+d\cdot\frac n2(n-1+2)=\frac{n+1}2[2a+(n+1-1)d]$