Could not see the difference between $a_{n}$ & $a_{n+1}$ is as given

Question

Could not see the difference between $a_{n}$ & $a_{n+1}$ is that what is given in the last line below ..... could anyone explain this for me please?

Answer 1

Suppose $n=5$. Ignore the square roots and reciprocals and just look at the $n(n+1)$ part.

The terms in $a_n$ start with $5\times 6$ and end with $9\times 10$. There are five terms.

The terms in $a_{n+1}$ start with $6\times 7$ and end with $11\times 12$. There are six terms.

Going from $5$ to $6$ you lose $5\times 6$ and gain $10\times 11$ and $11\times 12$. You gain one more term than you lose.

Does that help to see what is going on for the general $n$?

Answer 2

$$\require{cancel}$$ $$a_{n+1}-a_n=\cancel{\dfrac{1}{\sqrt{(n+1)(n+2)}}}+\cdots+\cancel{\dfrac{1}{\sqrt{(2n-1)(2n)}}}+\dfrac{1}{\sqrt{(2n)(2n+1)}}+\dfrac{1}{\sqrt{(2n+1)(2n+2)}}-\left(\dfrac{1}{\sqrt{(n)(n+1)}}+\cancel{\dfrac{1}{\sqrt{(n+1)(n+2)}}}+\cdots+\cancel{\dfrac{1}{\sqrt{(2n-1)(2n)}}}\right)$$