Let f be a polynomial function such that $$\log_{2}(f(x))=\left(\log_{2}\left(2+\frac{2}{3}+\frac{2}{9}+\ldots \ldots\right)\right) \cdot \log_{3}\left(1+\frac{f(x)}{f(1/x)}\right), x>0$$ and $f(6)=37$. Then $\sum_{n=1}^{10} f(n)$ is equal to ____.

JEE Main · 2026 · 08 Apr 2026 · Shift 2 · Sequences and Series

Let f be a polynomial function such that

lo g_{2} (f (x)) = (lo g_{2} (2 + \frac{2}{3} + \frac{2}{9} + \dots\dots)) \cdot lo g_{3} (1 + \frac{f ( x )}{f ( 1/ x )}), x > 0

and

f (6) = 37

. Then

\sum_{n = 1}^{10} f (n)

is equal to ____.

Answer: 395

The geometric series has first term

2

and ratio

\frac{1}{3}

, so

2 + \frac{2}{3} + \frac{2}{9} + \dots = \frac{2}{1 - \frac{1}{3}} = 3

. The equation becomes

lo g_{2} f (x) = lo g_{2} 3 \cdot lo g_{3} (1 + \frac{f ( x )}{f ( 1/ x )})

, and since

lo g_{2} 3 \cdot lo g_{3} y = lo g_{2} y

f (x) = 1 + \frac{f ( x )}{f ( 1/ x )} ⟹ f (x) f (1/ x) = f (x) + f (1/ x) .

A polynomial meeting this is

f (x) = 1 + x^{n}

, since

(1 + x^{n}) (1 + x^{- n}) = 2 + x^{n} + x^{- n} = (1 + x^{n}) + (1 + x^{- n})

. From

f (6) = 1 + 6^{n} = 37

we get

6^{n} = 36

, so

n = 2

and

f (x) = 1 + x^{2}

. Therefore

n = 1 \sum 10 f (n) = n = 1 \sum 10 (1 + n^{2}) = 10 + \frac{10 \cdot 11 \cdot 21}{6} = 10 + 385 = 395.

JEE Main · 2026 · 08 Apr 2026 · Shift 2 · Integer answer