Question

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

Answer 1

Hint: 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 \[{{S}_{n}}=\sum\limits_{n-1}^{n}{{{T}_{r}}}\].