Question

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

Answer 1

Hint:We know that $\sum\limits_{x = 1}^n {x = \dfrac{{n(n + 1)}}{2}}\,,\,\sum\limits_{x = 1}^n {{x^2}} = \dfrac{{n(n + 1)(2n + 1)}}{6}$.So we are given $1.2 + 2.3 + 3.4 + ....... + n(n + 1)$.So its general form is $n(n + 1)$.And we need to find $\sum\limits_{n = 1}^n n (n + 1)$ that is $\sum\limits_{n = 1}^n {{n^2}} + \sum\limits_{n = 1}^n n $.Using these formulas and prove that $1.2 + 2.3 + 3.4 + ....... + n(n + 1) = \dfrac{n}{3}(n + 1)(n + 2)$.

Complete step-by-step answer:
Here we are given that
$1.2 + 2.3 + 3.4 + ....... + n(n + 1)$
So we can write this as
$\sum\limits_{n = 1}^n n (n + 1)$
So upon expanding by putting $n = 1,2,3,4,.......,n$, we get
$1(1 + 1) + 2(2 + 1),,,,,,,,,, + n(n + 1)$
$1.2 + 2.3 + 3.4 + .......... + n(n + 1)$
So we get the value we need to find, so we can write that
$1.2 + 2.3 + 3.4 + ....... + n(n + 1)$ as $\sum\limits_{n = 1}^n n (n + 1)$
Now opening the bracket by multiplying the terms
$\sum\limits_{n = 1}^n {{n^2}} + \sum\limits_{n = 1}^n n $
So from (1) and (2), we get that
$\sum\limits_{i = 1}^n {i = \dfrac{{n(n + 1)}}{2}} $ and $\sum\limits_{i = 1}^n {{i^2}} = \dfrac{{n(n + 1)(2n + 1)}}{6}$
So we can expand by the above formula, we get that
So $\sum\limits_{n = 1}^n {{n^2}} + \sum\limits_{n = 1}^n n $
$ = \dfrac{{n(n + 1)(2n + 1)}}{6} + \dfrac{{n(n + 1)}}{2}$
As $\sum\limits_{x = 1}^n {x = \dfrac{{n(n + 1)}}{2}} $
     $\sum\limits_{x = 1}^n {{x^2}} = \dfrac{{n(n + 1)(2n + 1)}}{6}$
Now upon simplification, we can take $\dfrac{{n(n + 1)}}{2}$ common
We get
$ = \dfrac{{n(n + 1)}}{2}\left( {\dfrac{{2n + 1}}{3} + 1} \right)$
$ = \dfrac{{n(n + 1)}}{2}\left( {\dfrac{{2n + 1 + 3}}{3}} \right)$
$ = \dfrac{{n(n + 1)}}{2}\left( {\dfrac{{2n + 4}}{3}} \right)$
Now taking $2$ common from $2n + 4$, we get
$ = \dfrac{{2n(n + 1)(n + 2)}}{6}$
$ = \dfrac{{n(n + 1)(n + 2)}}{3}$
Hence proved.

Note:We know that when $1 + 2 + 3 + 4 + ....... + n$ is given, then its value is equal to $\dfrac{{n(n + 1)}}{2}$.This can be proved by using AP as we know that $1,2,3,4,.........,n$ are in AP with the common difference $d=1$ and first term $a=1$.So the sum of nth term is given by $ = \dfrac{n}{2}\left( {2a + (n - 1)d} \right)$,Here $a = 1,d = 1$
Sum$ = \dfrac{n}{2}\left( {2 + (n - 1)} \right)$$ = \dfrac{{n(n + 1)}}{2}$
Students should remember these formulas for solving the questions.
$\sum\limits_{i = 1}^n {i = 1 + 2 + 3 + 4 + ........ + n = \dfrac{{n(n + 1)}}{2}} $
$\sum\limits_{i = 1}^n {{i^2}} = {1^2} + {2^2} + {3^2} + ....... + {n^2} = \dfrac{{n(n + 1)(2n + 1)}}{6}$
$\sum\limits_{i = 1}^n {{i^3}} = {1^3} + {2^3} + {3^3} + ....... + {n^3} = {\left( {\dfrac{{n(n + 1)}}{2}} \right)^2}$.