Home
Class 12
MATHS
Given that u(n+1)=3un-2u(n-1), and u0=2 ...

Given that `u_(n+1)=3u_n-2u_(n-1),` and `u_0=2 ,u_(1)=3`, then prove that `u_n=2^(n)+1` for all positive integer of `n`

Text Solution

Verified by Experts

`because `U_(n+1)=3U_(n)-2U_(n-1)`
Step I `U_(1)=3=2+1=2^1+1` which is true for `n=1`.
Putting `n=1` in Eq. (i) we get
`U_(1+1)=3U_(1)-2U_(1-1)`
`rArr U_(2)=3u_(1)-2u_(0)=3.3-2.2=5=2^2+1` which is true for `n=2`
Therefore , the result is true for `n =1` and n=2`.
Step II Assume it is true for `n=k` , then it is also true for `n=k-`.
Then , `U_(k)=2^(k)+1`.....(ii)
and `u_(k-1)=2^(k-1)+1` ......(iii)
Step III Putting `n=k` in Eq. (i) we get
u_(k+1)=3u_(k)-2u_(k-1)`
`=3(2^k+1)-2(2^k-1+1)` [from Eqs. (ii) and (iii)]
`3.2^k+3-2.2^k-1-2=3.2^k+3-2^k-2`
`=(3-1)2^k+1=2.2^k+1=2^k+1+1`
This shows that the resutl is true for `n=k+1`. Hence , by the principle of mathematical induction the result is true for all `n in N`.
Promotional Banner

Similar Questions

Explore conceptually related problems

Prove that 2^(n) gt n for all positive integers n.

Prove that n^(2)-n is divisible by 2 for every positive integer n.

If A= [[0,1,0],[1,0,0],[0,0,1]] , then A^(2n)= where n is a positive integer.

If n is a positive integer, then (1+i)^(n)+(1-i)^(n)=

Prove that the Binomial theorem (a + b)^(n) = ""^(n)C_(0)a^(n) + ""^(n)C_(1)a^(n - 1)b + ""^(n)C_(2)a^(n - 2)b^(2) + .. ""^(n)C_(n)b^(n) for any positive integer 'n'.

Let A = [(1,0,0,),(2,1,0),(3,2,1)] . "If u_(1) and u_(2) are column matrices such that Au_(1) = [(1),(0),(0)] and Au_(2) = [(0),(1),(0)] then u_(1) +u_(2) is equal to :

Show that the middle term in the expansion of (1+x)^(2n) is 1.3.5…(2n-1)/n! 2n.x^n , where n is a positive integer.

If U_(n) = int_0^(pi/4) tan^(n) x dx then u_(n)+u_(n-2) =

Let u_(1)=1,u_2=2,u_(3)=(7)/(2)and u_(n+3)=3u_(n+2)-((3)/(2))u_(n+1)-u_(n) . Use the principle of mathematical induction to show that u_(n)=(1)/(3)[2^(n)+((1+sqrt(3))/(2))^n+((1-sqrt(3))/(2))^n]forall n ge 1 .