Question

If $P\left(n\right):2\cdot 4^{2n+1}+3^{3n+1}$ is divisible by $\lambda$ for all $n\in N,$ then which of the following is/are correct?
(use mathematical induction property)

• A. $\lambda -11=0$
• B. $\lambda -7=0$
• C. unit place of $2^{\lambda }$ is $8$
• D. unit place of $3^{\lambda }$ is $7$

Answer 1

Answer: (A), (C), (D)

unit place of $3^{\lambda }$ is $7$

Given :
$P\left(n\right):2\cdot 4^{2n+1}+3^{3n+1}$ is divisible by $\lambda$
From property of mathematical induction : $P\left(n\right)$ is true for $n=1,2,3,\cdots$
Now,
$P\left(1\right)=2\cdot 64+81=209$
$P\left(2\right)=2\cdot 4^5+3^7=4235$
So, value of $\lambda =\text{H.C.F. of}(209,4235)=11$
Let $P\left(n\right):2\cdot 4^{2n+1}+3^{3n+1}$ is divisible by $11$ is true for $k=n.$
check at $k=n+1,$
$P(n+1):2\cdot 4^{2n+3}+3^{3n+4}=16(2\cdot 4^{2n+1}+3^{3n+1})+11\cdot 3^{3n+1}$ which is divisible by $11$ is true
$[\because 2\cdot 4^{2n+1}+3^{3n+1}\text{is divisible by}11]$
So, $P(n+1)$ is also true.