IIT JEE 1982 Maths - 'Adapted' - Subjective to Multi-Correct Q10

$\frac {1}{ \sqrt {n-1} + \sqrt {n} } = \frac {1}{ \sqrt {n-1} + \sqrt {n} } × \frac { \sqrt {n} - \sqrt {n-1} }{ \sqrt {n} - \sqrt {n-1} } = \frac { \sqrt {n} - \sqrt {n-1} }{n - (n-1)} = \sqrt {n} - \sqrt {n-1}$

Applying this for all terms in f(n), we get the following telescopic sum:

$f(n) = ( \sqrt {2} - 1) + ( \sqrt {3} - \sqrt {2} ) + ( \sqrt {4} - \sqrt {3} ) + ... + ( \sqrt {n - 1} - \sqrt {n - 2} ) + ( \sqrt {n} - \sqrt {n - 1} ) = \sqrt {n} - 1$

Hence, all of the statements are false:

$f(10) = \sqrt {10} - 1 ≠ 2$

$f(15) = \sqrt {15} - 1 ≠ 3$

$f(20) = \sqrt {20} - 1 ≠ 4$

$f(25) = \sqrt {25} - 1 = 5 - 1 = 4 ≠ 5$

Therefore, our answer should be:

$\boxed {9999}$

By conjugate multiplication, we can see that $f(n)$ comes out after cutting each other that $f(n)$ = $\sqrt{n}$ - 1

Hence it satisfies none of A,B,C,D Hence answer 9999.

$\begin{aligned} f(n) & = \frac 1{\sqrt 1+\sqrt 2} + \frac 1{\sqrt 2+\sqrt 3} + \frac 1{\sqrt 3+\sqrt 4} + ... + \frac 1{\sqrt {n-1}+\sqrt n} \\ & = \sum_{k=2}^n \frac 1{\sqrt{k-1}+\sqrt k} \\ & = \sum_{k=2}^n \frac {\sqrt k - \sqrt {k-1}}{(\sqrt k + \sqrt {k-1})(\sqrt k - \sqrt {k-1})} \\ & = \sum_{k=2}^n \frac {\sqrt k - \sqrt {k-1}}{k-(k-1)} \\ & = \sum_{k=2}^n \left( \sqrt k - \sqrt {k-1} \right) \\ & = \sum_{k=2}^n \sqrt k - \sum_{k=1}^{n-1} \sqrt k \\ & = \sqrt n - 1 \end{aligned}$

$\implies \begin{cases} f(10) = \sqrt {10} - 1 \color{#D61F06} \ne 2 \\ f(15) = \sqrt {15} - 1 \color{#D61F06} \ne 3 \\ f(20) = \sqrt {20} - 1 \color{#D61F06} \ne 4 \\ f(25) = \sqrt {25} - 1 \color{#D61F06} \ne 5 \end{cases} \implies$ the answer is $\boxed{9999}$ .