Question

# Prove that $1.2 + 2.3 + .......n(n + 1) = \dfrac{{n(n + 1)(n + 2)}}{3}$​ in mathematical induction.

Hint: Here we are given a typical mathematical induction problem to deal with. First we need to check if the expression is true for $n = 1$, then we need to consider that it is true for $n = k\;$and then need to check if it’s true for $n = k + 1$, if true for $n = k + 1$ we can say that the equations holds true.

We shall prove the result by principle of mathematical induction.
$p(n) = 1.2 + 2.3 + .......n(n + 1) = \dfrac{{n(n + 1)(n + 2)}}{3}$
First on checking for $n = 1$,
We have,
LHS: $1.2 = 2$
RHS: $\dfrac{{n(n + 1)(n + 2)}}{3}$
On substituting the value of n we get,
$= \dfrac{{1(1 + 1)(1 + 2)}}{3} \\ = \dfrac{{1.2.3}}{3} \\ = 2. \\$
As ${\text{LHS = RHS}}$
Hence p(n) is true for n=1.
Now,
Let us assume the result is true for $n = k\;$ i.e.,
$p(k) = 1.2 + 2.3 + .....k\left( {k + 1} \right) = \dfrac{{k \times \left( {k + 1} \right) \times \left( {k + 2} \right)}}{3}\;$
We shall prove that p(n) is true for $n = k + 1$.
that is, to prove $p(k + 1) = 1.2 + 2.3..... + k\left( {k + 1} \right) + \left( {k + 1} \right)\left( {k + 2} \right) = \dfrac{{\left( {k + 1} \right)\left( {k + 2} \right)\left( {k + 3} \right)}}{3}$
Consider LHS: $1.2 + 2.3..... + k\left( {k + 1} \right) + \left( {k + 1} \right)\left( {k + 2} \right)$
As, p(k) is true we have $p(k) = 1.2 + 2.3 + .....k\left( {k + 1} \right) = \dfrac{{k \times \left( {k + 1} \right) \times \left( {k + 2} \right)}}{3}\;$,
Hence on substituting the above value we get,
$= \dfrac{{k \times \left( {k + 1} \right) \times \left( {k + 2} \right)}}{3} + \left( {k + 1} \right)\left( {k + 2} \right)$
Taking common, $\left( {k + 1} \right)\left( {k + 2} \right)$, we get,
$= \left( {k + 1} \right)\left( {k + 2} \right)[\dfrac{k}{3} + 1]$
On simplifying we get,
$= \left( {k + 1} \right)\left( {k + 2} \right)[\dfrac{{k + 3}}{3}]$
$= [\dfrac{{\left( {k + 1} \right)\left( {k + 2} \right)(k + 3)}}{3}]$​
=RHS.
Hence the p(n) holds for $n = k + 1$.
As, p(n) is true for $n = 1$, $n = k\;$and $n = k + 1$, hence we can say that ${\text{p(n)}}$ is true$\forall n \in N$.
Hence, proved.

Note: Mathematical induction is a mathematical proof technique. It is essentially used to prove that a statement $P\left( n \right)$holds for every natural number $n\; = \;0,\;1,\;2,\;3,\;$. . . ; that is, the overall statement is a sequence of infinitely many cases $P\left( 0 \right),\;P\left( 1 \right),\;P\left( 2 \right),\;P\left( 3 \right),$. . . . . We generally assume that $P\left( n \right)$is true for $n = k\;$and using this we prove that $P\left( n \right)$ is true for $n = k + 1$.