Question

$a+(a+d)+(a+2d)+....+(a+(n-1\left)d\right)=\frac{n}{2}[2a+(n-1\left)d\right]$

Answer 1

Let P(n) : $a+(a+d)+(a+2d)+......+(a+(n-1\left)d\right)=\frac{n}{2}[2a+(n-1\left)d\right]$

For n = 1

$a=\frac{1}{2}[2a+(1-1\left)d\right]$

$a=a$

$\Rightarrow$ P(n) is true for n = 1

Let P(n) is true for n = k, so

$a+(a+d)+(a+2d)+...+(a+(k-1\left)d\right)=\frac{k}{2}[2a+(k-1\left)d\right]$ .......(1)

We have to show that,

$a+(a+d)+(a+2d)+......+(a+(k-1\left)d\right)+(a+(k\left)d\right)=\frac{(k+1)}{2}[2a+kd]$

Now,

$\{a+(a+d)+(a+2d)+....+(a+(k-1)d\left)\right\}+(a+kd)$

$=\frac{k}{2}[2a+(k-1\left)d\right]+(a+kd)$

[Using equation (1)]

$=\frac{2ka+k(k-1)d+2(a+kd)}{2}$

$=\frac{2ka+k^{2}d-kd+2a+2kd}{2}$

$=\frac{2ka+2a+k^{2}d+kd}{2}$

$=\frac{2a(k+1)+d(k^2+k)}{2}$

$=\frac{(k+1)}{2}[2a+kd]$

$\Rightarrow$ P(n) is true for n = k + 1

$\Rightarrow$ P(n) is true for all n $ϵ$ N by PMI.