Question

If $1+\dfrac{1+2}{2}+\dfrac{1+2+3}{3}+...$ to n terms is S, the S is equal to?

Answer 1

In this problem we have to find the total sum of the given sequence, ${{S}_{n}}$. Here we can use the arithmetic sequence formulas. We can first find the ${{r}^{th}}$ term of the given sequence. We know that the formula to find the sum of the given sequence is ${{S}_{n}}=\sum\limits_{n-1}^{n}{{{T}_{r}}}$. Here we can substitute the ${{r}^{th}}$ term and simplify the summation to get the required answer.

Complete step by step answer:
Here we have to find the total sum of the given sequence.
We can see that the given sequence is,
${{S}_{n}}=1+\dfrac{1+2}{2}+\dfrac{1+2+3}{3}+.....$
We can now find the ${{r}^{th}}$ term of the given sequence.
We know that the formula to find the ${{r}^{th}}$term of the given sequence is,
${{T}_{r}}=\dfrac{r\left( r+1 \right)}{2r}=\dfrac{r+1}{2}$
We know that the formula to find the sum of the given sequence is,
$\Rightarrow {{S}_{n}}=\sum\limits_{n=1}^{n}{{{T}_{r}}}$.
We can now substitute the ${{r}^{th}}$ term and simplify the summation, we get
$\Rightarrow {{S}_{n}}=\sum\limits_{n=1}^{n}{\dfrac{r+1}{2}}$
We can now simplify the above step and write it as,
$\Rightarrow {{S}_{n}}=\dfrac{1}{2}\left( \sum{r}+\sum{1} \right)=\dfrac{1}{2}\left[ \dfrac{n\left( n+1 \right)}{2}+n \right]$
We can further simplify the above step, we get
$\Rightarrow {{S}_{n}}=\dfrac{1}{2}\left[ \dfrac{n\left( n+1 \right)}{2}+n \right]=\dfrac{2n+n\left( n+1 \right)}{4}=\dfrac{n\left( n+3 \right)}{4}$
Therefore, the sum of the given sequence is ${{S}_{n}}=\dfrac{n\left( n+3 \right)}{4}$.

Note: We should always remember the arithmetic general formulas such as the formula to find the ${{r}^{th}}$ term and the formula to find the sum of the sequence. We should remember that the formula to find the ${{r}^{th}}$ term is ${{T}_{r}}=\dfrac{r\left( r+1 \right)}{2r}=\dfrac{r+1}{2}$ and the formula to find the sum of the given sequence with the ${{r}^{th}}$ is ${{S}_{n}}=\sum\limits_{n-1}^{n}{{{T}_{r}}}$.