The sum of the series $$\dfrac{1}{1+\sqrt{2}} + \dfrac{1}{\sqrt{2}+\sqrt{3}} + \dfrac{1}{\sqrt{3}+\sqrt{4}} + \ldots$$ upto $$15$$ terms is

Consider the general term of the given series
$$T_k \;=\;\frac{1}{\sqrt{k}\;+\;\sqrt{k+1}}$$ for $$k = 1,2,3,\dots,15$$.

Step 1 — Rationalise the denominator.
Multiply numerator and denominator by $$\sqrt{k+1}-\sqrt{k}$$:
$$T_k \;=\;\frac{1}{\sqrt{k}+\sqrt{k+1}}\;\cdot\;\frac{\sqrt{k+1}-\sqrt{k}}{\sqrt{k+1}-\sqrt{k}}$$
$$=\;\frac{\sqrt{k+1}-\sqrt{k}}{(\sqrt{k+1})^{2}-(\sqrt{k})^{2}}$$
$$=\;\frac{\sqrt{k+1}-\sqrt{k}}{(k+1)-k}$$
$$=\;\sqrt{k+1}-\sqrt{k}.$$

Step 2 — Write the required sum $$S_{15}$$ using the simplified term.
$$S_{15}\;=\;\sum_{k=1}^{15}\left(\sqrt{k+1}-\sqrt{k}\right).$$

Step 3 — Observe the telescoping nature.
$$\bigl(\sqrt{2}-\sqrt{1}\bigr)\;+\;\bigl(\sqrt{3}-\sqrt{2}\bigr)\;+\;\dots\;+\;\bigl(\sqrt{16}-\sqrt{15}\bigr)$$
All intermediate terms cancel pair-wise, leaving only the first negative and the last positive term:

$$S_{15}\;=\;\sqrt{16}-\sqrt{1}\;=\;4-1\;=\;3.$$

Hence the sum of the first 15 terms is $$3$$.

Option C which is: $$3$$