Q. 4

# If P(n): 2 × 4^{2n + 1} + 3^{3n + 1} is divisible by λ for all n ∈ N is true, then find the value of λ.

Answer :

for n=1,

2×4^{2×1+1} + 3^{3×1+1}=2×4^{3}+3^{4}

= 2×64+81

= 128+81

= 209

For n=2,

2×4^{2×2+1} + 3^{3×2+1} = 2×4^{5}+3^{7}

= 2×1024+2187

= 2048+2187

= 4235

Now, the H.C.F of 209 and 4235 is 11.

Hence, λ=11.

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