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Question

Prove that  2.7n+3.5n5 is divisible by 24 true for all natural numbers.


Solution

Let p (n) be the statement given by
P(n): 2.7n+3.5n5 is divisible by 24
Then p(1): 2×71+3×515=14+155=24 which is divisible by 24
p(1) is true
Let p(m): 2.7m+3.5m5 is divisible by 24 be true
 2.7m+3.5m5=24λ; λϵN
 3.5m=24λ+52.7m . . . . . (1)
Now
p(m+1):2.7m+1+3.5m+15
=2.7m+1+(3.5m)55
=2.7m+1+(24 λ+52.7m)55  (from(1))
=2.7m+1+120λ+2510.7m5
=(2.7m+110.7m)+120λ+20
=(2×7λ 7m10×7m)+120λ+244
=(1410)7m4+24(5λ+1)
=4(7m1)+24(5λ+1)
=4×6μ+24(5λ+1)( 7m1 is a multiple of 6 for all m ϵ N   7m1=6μ, μ ϵ N)
=24 (μ+5λ+1) which is divisible by 24
p(m+1) is true.
Hence by principle of mathematical induction p(n) is true for all n ϵ N

All India Test Series

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