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BITSAT 2023 Shift 2 Solved Paper

Section: Mathematics

Question : 102 of 130

Marks: +1, -0

n

2 \cdot 4^{2n+1}+3^{3n+1}

Solution: 👈: Video Solution

Let

P(n)=2 \cdot 4^{2n+1}+3^{3n+1}

Then

P(1)=2 \cdot 4^3+3^4=209

, which is divisible by 11 but not divisible by 2,7 or 27 .

Further, let

P(k)=2 \cdot 4^{2k+1}+3^{3k+1}

is divisible by 11 , that is,

2 \cdot 4^{2k+1}+3^{3k+1}=11 q

for some integer

q

. Now

P(k+1)\;=2 \cdot 4^{2k+3}+3^{3k+4}

\;=2 \cdot 4^{2k+1} \cdot 4^2+3^{3k+1} \cdot 3^3

\;=16 \cdot 2 \cdot 4^{2k+1}+27 \cdot 3^{3k+1}

\;=16 \cdot 2 \cdot 4^{2k+1}+(16+11) \cdot 3^{3k+1}

\;=16 \left[2 \cdot 4^{2k+1}+3^{3k+1}\right] +11 \cdot 3^{3k+1}

\;=16 \cdot 1 lq+11 \cdot 3^{3k+1}

\;=11 (16 q+3^{3k+1}) =11 m

where

m=16 q+3^{3k+1}

is another integer.

\therefore P(k+1)

is divisible by 11 .

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