First term of the sequence be 1 and \[{(n + 1)^{th}}\] term is obtained by adding (n+1) to the \[{n^{th}}\] term. Find the sequence upto \[{6^{th}}\] term.

Hint: Try to formulate a general equation for getting each term of the sequence. After doing that start by finding the second term (as the first term is already given). Eventually try to find the successor of each term by using the value you already have. For example if you found the second term use that to find the third term, proceed in this way till you get the \[{6^{th}}\] term.
Complete Step by step Solution:
First of all let's try to find the generalized equation, it is given that we can get \[{(n + 1)^{th}}\] term by adding (n+1) to the \[{n^{th}}\] term.
Let us denote \[{n^{th}}\] term by \[{a_n}\] then \[{(n + 1)^{th}}\] term will be \[{a_{n + 1}}\]
Now we can write
\[{a_{n + 1}} = {a_n} + (n + 1)\]
The above mentioned equation is a general equation of all terms
We already have the first term lets try to find the second term i.e., \[{a_2}\]
For getting \[{a_2}\] , Put \[n = 1\]
\[\begin{array}{l}
\therefore {a_{n + 1}} = {a_n} + (n + 1)\\
 \Rightarrow {a_{1 + 1}} = {a_1} + (1 + 1)\\
 \Rightarrow {a_2} = {a_1} + 2\\
 \Rightarrow {a_2} = 3
\end{array}\]
Now using \[{a_2}\] let's try to find \[{a_3}\]
\[\begin{array}{l}
\therefore {a_{n + 1}} = {a_n} + (n + 1)\\
 \Rightarrow {a_{2 + 1}} = {a_2} + (2 + 1)\\
 \Rightarrow {a_2} = {a_2} + 3\\
 \Rightarrow {a_3} = 6
\end{array}\]
Similarly \[{a_4}\] will be
\[\begin{array}{l}
\therefore {a_{n + 1}} = {a_n} + (n + 1)\\
 \Rightarrow {a_{3 + 1}} = {a_3} + (3 + 1)\\
 \Rightarrow {a_4} = {a_3} + 4\\
 \Rightarrow {a_4} = 6 + 4\\
 \Rightarrow {a_4} = 10
\end{array}\]
Again \[{a_5}\] will be
\[\begin{array}{l}
\therefore {a_{n + 1}} = {a_n} + (n + 1)\\
 \Rightarrow {a_{4 + 1}} = {a_4} + (4 + 1)\\
 \Rightarrow {a_5} = {a_4} + 5\\
 \Rightarrow {a_5} = 10 + 5\\
 \Rightarrow {a_5} = 15
\end{array}\]
Now for \[{a_6}\] we will use \[{a_5}\]
\[\begin{array}{l}
\therefore {a_{n + 1}} = {a_n} + (n + 1)\\
 \Rightarrow {a_{5 + 1}} = {a_4} + (5 + 1)\\
 \Rightarrow {a_6} = {a_4} + 6\\
 \Rightarrow {a_6} = 15 + 6\\
 \Rightarrow {a_6} = 21
\end{array}\]
Hence the sequence is
1,3,6,10,15,21,....................

Note: Always use the predecessor of the term while trying to find a new one and make sure to change the value of n in (n+1) students often forget to change it which results in an incorrect sequence.