1.

For all `n in N, 2^(4n)-15n-1` is divisible byA. 125B. 225C. 450D. None of these

Answer» Correct Answer - B
Let `P(n) : 2^(4n) - 15 n - 1`
Put n = 2
`P(2)=2^(8)-30-1=225` which is divisible by 225.
Let us assume,
P(n) is true for n = k is P(K) : `2^(4k)-15k-1` is divisible by 225.
`rArr 2^(4k)-15k-1=225 lambda, lambda in R, k in N" "...(i)`
To prove for n = k + 1
Consider
`2^(4k+4)-15k-15-1=2^(4k).2^(4)-15k-16`
`=2^(4)[225 lambda + 1 + 15k]-15k-16" "("from (i)")`
`=2^(4).225 lambda .2^(4)+15.2^(4).k-15k-16`
`=25.225 lambda + 225 k`
`=225[2^(4) lambda + k]`
= 225 r where
`r = 2^(4)lambda + k` is a constant
Hence, `2^(4n)-15n - 1` is divisible by 225.


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