[JEE Advanced 1985] prove by induction that 2 7 n 3 5 n 5 is divisible by
Prove by induction that $2\cdot7^n+3\cdot5^n-5$ is divisible by $24$ for all $n>0$.
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Answer: Proved.
Base $n=1$: $14+15-5=24$. The step adds $2\cdot6\cdot7^n+3\cdot4\cdot5^n=12(7^n+5^n)$, and $7^n+5^n$ is even, so the increment is divisible by $24$.
JEE Advanced 1985 · Binomial Theorem — verified solution by the Vidaara Team.
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