[JEE Advanced 1990] Prove that $\dfrac{n^7}7+\dfrac{n^5}5+\dfrac{2n^3}3-\dfrac n{105}$ is an integer for every positive integer $n$.
Prove that $\dfrac{n^7}7+\dfrac{n^5}5+\dfrac{2n^3}3-\dfrac n{105}$ is an integer for every positive integer $n$.
1 Answer
VAVidaara Admin
✓ Vidaara Team
✓ Accepted
· 1mo ago
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Answer: Proved.
Combining over $105$: $\frac{15n^7+21n^5+70n^3-n}{105}$; the numerator is divisible by $3,5,7$ (checked via Fermat/parity), hence by $105$.
JEE Advanced 1990 · Binomial Theorem — verified solution by the Vidaara Team.
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