Question

Evaluate: $1.2+2.3+3.4+\dots +n(n+1)=\frac{n}{3}(n+1)(n+2)$

Answer 1

Given,

to evaluate

$1.2+2.3+3.4+....+n(n+1)=\frac{n}{3}(n+1)(n+2)$

Let,

$P\left(n\right):1.2+2.3+3.4+.....+n(n+1)=\frac{n(n+1)(n+2)}{3}$

for $n=1$, we get

L.H.S $=n(n+1)=1(1+1)=1\left(2\right)=2$.

R.H,S $=\frac{n(n+1)(n+2)}{3}=\frac{i(1+1)(1+2)}{3}\Rightarrow \frac{2\times 3}{3}=2$

$\therefore$ L.H.S=R.H.S

So, $P\left(n\right)$ is true for $n=1$

Now, let us assume $P\left(n\right)=P\left(k\right)$

we get,

$P\left(K\right):1.2+2.3+3.4+......+k(k+1)-----\left(1\right)$

$=\frac{k(k+1)(k+2)}{3}$

Let us prove that $(k+1)$ will be true for $P\left(k\right)$ so, we get

$n=k+1$ from eq $\left(1\right)$

$L\Rightarrow 1.2+2.3+3.4+....+(k+1)\left(\right(k+1)+1)=\frac{(k+1)\left(\right(k+1)+1)\left(\right(k+1)+2)}{3}$

$L\Rightarrow 1.2+2.3+3.4+.....+(k+1)(k+2)=\frac{(k+1)(k+2)}{3}$

$\Rightarrow 1.2+2.3+3.4+....+k(k+1)+(k+1)(k+2)=\frac{(k+1)(k+2)(k+3)}{3}$

Now, let us consider $P\left(k\right)3$

we get,

$1.2+2.3+3.4+....+k.(k+1)=\frac{(k+1)(k+2)}{3}$

Now let us add $(k+1)(k+2)$ b/s we get,

$1.2+2.3+3.4+....+k(k+1)+(k+1)(k+2)\Rightarrow \frac{(k+1)(k+2)}{3}+(k+1).(k+2)$

Let us consider R.H.S, we get,

$\Rightarrow \frac{k(k+1)(k+2)+3\left[\right(k+1\left)\right(k+2\left)\right]}{3}$

$\Rightarrow \frac{k(k+1)(k+2)+3(k+1)(k+2)}{3}$

(by taking common)

$\Rightarrow \frac{(k+3)\left[\right(k+1\left)\right(k+2\left)\right]}{3}$

$\Rightarrow \frac{(k+1)(k+2)(k+3)}{3}$

Therefore, we get $P\left(k\right)$ as

$1.2+2.3+3.4+.....+k.(k+1)+(k+1)(k+2)=\frac{(k+1)(k+2)(k+3)}{3}$

$\therefore$ for $P\left(k\right)=0$ we get $P\left(k\right)=0$ and $\forall P\left(k\right)=n$ we get $P(k+1)=n$

So, we can say that $P(k+1)$ will be true for $P\left(k\right)$ true form.

And from the principle of mathematics induction, $P\left(n\right)$ is all true for ${}^{\prime }{n}^{\prime }$, when ${}^{\prime }{n}^{\prime }$ is a natural number.