Question

Prove that ${1+(1+2)+(1+2+3)+ ... +(1+2+3+4+...+n) }$ is equal to $\orange{\dfrac{n(n+1)(n+2)}{6}}$where n is the no. of terms in first n natural numbers.​

Answer 1

$\large\underline{\sf{Solution-}}$

Consider,

$\rm \: 1 + (1 + 2) + (1 + 2 + 3) + - - - + (1 + 2 + 3 + - - - + n)$

Let assume that,

$\rm \:S_n = 1 + (1 + 2) + (1 + 2 + 3) + - - - + (1 + 2 + 3 + - - - + n)$

So, nth term of the series is

$\rm \: T_n = 1 + 2 + 3 + - - + n$

$\rm \: T_n = \displaystyle\sum_{k=1}^n\rm k$

$\rm \: T_n = \dfrac{n(n + 1)}{2}$

$\rm \: T_n = \dfrac{1}{2}( {n}^{2} + n)$

So, Sum of n terms of series is given by

$\rm \: S_n = \displaystyle\sum_{k=1}^n\rm T_k$

$\rm \: S_n = \dfrac{1}{2} \displaystyle\sum_{k=1}^n\rm ( {k}^{2} + k)$

$\rm \: S_n = \dfrac{1}{2}\bigg( \displaystyle\sum_{k=1}^n\rm {k}^{2} + \displaystyle\sum_{k=1}^n\rm k\bigg)$

$\rm \: S_n = \dfrac{1}{2}\bigg(\dfrac{n(n + 1)}{2} + \dfrac{n(n + 1)(2n + 1)}{6} \bigg)$

$\rm \: S_n = \dfrac{1}{2} \times \dfrac{n(n + 1)}{2}\bigg(1 + \dfrac{2n + 1}{3} \bigg)$

$\rm \: S_n = \dfrac{1}{2} \times \dfrac{n(n + 1)}{2}\bigg(\dfrac{3 + 2n + 1}{3} \bigg)$

$\rm \: S_n = \dfrac{1}{2} \times \dfrac{n(n + 1)}{2}\bigg(\dfrac{2n + 4}{3} \bigg)$

$\rm \: S_n = \dfrac{1}{2} \times \dfrac{n(n + 1)}{2}\bigg(\dfrac{2(n + 2)}{3} \bigg)$

$\rm\implies \:S_n = \dfrac{n(n + 1)(n + 2)}{6}$

Hence, Proved

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Formula Used :-

$\boxed{\sf{ \:\displaystyle\sum_{k=1}^n\rm k \: = \: \frac{n(n + 1)}{2} \: }}$

$\boxed{\sf{ \:\displaystyle\sum_{k=1}^n\rm {k}^{2} \: = \: \frac{n(n + 1)(2n + 1)}{6} \: }}$

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Additional Information

$\boxed{\sf{ \:\displaystyle\sum_{k=1}^n\rm {k}^{3} \: = \: {\bigg[\dfrac{n(n + 1)}{2} \bigg]}^{2} \: }}$

$\boxed{\sf{ \:\displaystyle\sum_{k=1}^n\rm 1 \: = \: n \: }}$

Answer 2

Step-by-step explanation:

Answer:-

In the series $\sf 1+(1+2)+(1+2+3) \ldots \ldots$ n th term is the sum of numbers from 1 ton. So, the n th term is $\tt\dfrac{n(n+1)}{2}$

So, we need to find the sum of a series whose n th term is $\tt\frac{n(n+1)}{2}=\frac{n^{2}+n}{2}$

Therefore:-

$\[ \begin{aligned} \tt \color{red}\sum \frac{n^{2}}{2} &\tt \color{red}=\frac{n(n+1)(2 n+1)}{6 \times 2}=\frac{n(n+1)(2 n+1)}{12} \\ \\ \tt \color{blue}\sum \frac{n}{2} &\tt \color{blue}=\frac{n(n+1)}{2 \times 2}=\frac{n(n+1)}{4} \\ \\\tt \color{orange} \sum \frac{n(n+1)}{2} &\tt \color{orange}=\sum \frac{n^{2}}{2}+\sum \frac{n}{2} \\ \\ &\tt \color{darkred}=\frac{n(n+1)(2 n+1)}{12}+\frac{n(n+1)}{4 } \\ \\ &\tt \color{green}=\frac{n(n+1)}{12}[(2 n+1)+3] \\ \\ &\tt \color{purple}=\frac{n(n+1)}{12}(2 n+4) \\ \\ & \boxed{\tt \pink{=\frac{n(n+1)(n+2)}{6} } }\\ \\ & \color{lime} \mathbb{\therefore \: HENCE, \: \: \: FINAL \: \: \: ANSWER}\end{aligned} \]$