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Question

# If n∈N , then 72n+23n−3.3n−1 is always divisible by

A
25
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B
None of these
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C
35
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D
45
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Solution

## The correct option is A 2572n+23n−3×3n−1=49n+23n23×3n3⇒49n+23n+3n24=49n+(23×3)n24=49n+24n24⇒49n+24n−149≡24(mod25)49n≡24n(mod25)49n≡25×m+24nm∈ 49n+24n−1=25×m+24n+24n−1 49n+24n−1=25m+(24+1)×24n−1⇒25(m+24n−1)≡0mol25[nϵN]So, option (A) is correctIInd method-we can prove this by mathimationinduction.for n = 1. 72n+23n−3.3n−149+1×1=50it is divisible by 1, 2, 5, 10, 25 ,50so, option (c), (D), us discordAns Now, if it is true for ∀n thenoption (A) is correct.So, let the result be one true for n=k i.e72k+23(k−1),3k−1 is a multiple of 25Now, we need to prove that the result is also true for n=k+1, that is72k+2+23k.3k is a multiple of 25 ⇒72k+2+23k.3k=72.72k+23.23(k−1).3.3k−1=49.72k+49.23(k−1).3k−1−2523(k−1).3k−1=49(72k+23(k−1).3k−1)−25.23(k−1).3k−1by owr assumption72k+23(k−1).3k−1 is a multiple of 25.⇒49(72k+23(k−1).3k−1) is amultiple of 25and also 25.23(k−1).3k−1 is a multiple of 25⇒49(72k+23(k−1).3k−1)−25.23−(k−1).3k−1is a multiple of 25.72k+2+23k.3k is a multiple of 25.∴ The result is true for n=k+1 hence,it is true ∀nSo option (A) is correct.

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