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

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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.

Complete step-by-step answer:
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 \[n = k + 1\].