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Question:

A function f(x) satisfies f(1) = 3600, and f(1) + f(2) + ... + f(n) = n²f(n), for all positive integers n > 1. What is the value of f(9)?

Explanation:

f(1) + f(2) + f (3) + … + f(n −1) + f(n) = n²f(n)  ... (i)

Similarly, f(1) + f(2) + f (3) + … + f(n − 1) = (n − 1)² f(n −1)  ... (ii)

⇒ f(n) = n² f(n) – (n – 1)²f(n − 1) ... (i) – (ii)

⇒ (n² – 1)f(n) = (n – 1)²f(n – 1)

⇒ f(n) = $\frac{{(\mathrm{n}-1)}^2\mathrm{f}(\mathrm{n}-1)}{({\mathrm{n}}^2-1)}$

⇒ f(n) = $\frac{(\mathrm{n}-1)\mathrm{f}(\mathrm{n}-1)}{(\mathrm{n}+1)}$

Now, putting

n = 2 we get f(2) = $\frac{1}{3}\times \mathrm{f}\left(1\right)$
n = 3 we get f(2) = $\frac{2}{4}\mathrm{f}\left(2\right)$ = $\frac{2}{4}\times \frac{1}{3}\times \mathrm{f}\left(1\right)$
...
n = 9 we get f(9) = $\frac{8}{10}\mathrm{f}\left(8\right)$ = $\frac{8}{10}\times \frac{7}{9}\times \frac{6}{8}\times \frac{5}{7}\times \frac{4}{6}\times \frac{3}{5}\times \frac{2}{4}\times \frac{1}{3}\times \mathrm{f}\left(1\right)$

∴  f(9) = $\frac{8}{10}\times \frac{7}{9}\times \frac{6}{8}\times \frac{5}{7}\times \frac{4}{6}\times \frac{3}{5}\times \frac{2}{4}\times \frac{1}{3}\times 3600$

= $\frac{2}{10\times 9}\times 3600$ = 80

Hence, option (a).