BITSAT Mathematics Question (2023) — Solution

Step-by-step solution

Let P(n)=2 4^ 2 n+1 +3^ 3 n+1 Then P(1)=2.4^3+3^4=209, which is divisible by 11 but not divisible by 2,7 or 27 . Further, let P(k)=2.4^ 2 k+1 +3^ 3 k+1 is divisible by 11 , that is, 2.4^ 2 k+1 +3^ 3 k+1 =11 q for some integer q. Now aligned P ( k +1) & =2 4^ 2 k +3 +3^ 3 k +4 \\ & =2 4^ 2 k +1 4^2+3^ 3 k +1 3^3 \\ & =16 2 4^ 2 k +1 +27 3^ 3 k +1 \\ & =16 2 4^ 2 k +1 +(16+11) 3^ 3 k +1 \\ & =16 [2 4^ 2 k +1 +3^ 3 k +1 ]+11 3^ 3 k +1 \\ & =16 11 q +11 3^ 3 k +1 \\ & =11 (16 q +3^ 3 k +1 )=11 ~m aligned where m=16 q+3^ 3 k+1 is another integer. P ( k +1) is divisible by 11 .