How many numbers between 99 to 999 which are divisible by 11.

Explanation

`because u_(n)=(1)/(3)[2^(n)+((1+sqrt(3))/(2))^n+((1-sqrt(3))/(2))^n]` ......(i)
Step I For `n=1, u_(1)=(1)/(3)[2^1+((1+sqrt(3))/(2))^1+((1-sqrt(3))/(2))^1]=(1)/(3)[2+1]=1`
Which is true for `n=` and for `n=2`,
`u_(2)=(1)/(3)[2^2+((1+sqrt(3))/(2))^2+((1-sqrt(3))/(2))^2]`
`=(1)/(3) [4+((4+2sqrt(3))/(4))+((4-2sqrt(3))/(4))]=(1)/(3)[6]=2`
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-1, k-2`
`therefore u_(k)=(1)/(3)[2^k+((1+sqrt(3))/(2))^k+((1-sqrt(3))/2)^k]`....(ii)
`u_(k-1)=(1)/(3)[2^(k=-1)+((1+sqrt(3))/(2))^(k-1)+((1-sqrt(3))/(2))^(k-1)]`.....(iii)
`u_(k-2)=(1)/(3)[2^(k-2)+((1+sqrt(3))/(2))^(k-2)+((1-sqrt(3))/(2))^(k-2)]`.....(iv)
Step III Given that , `u_(n+3)=3u_(n+2)-((3)/(2))u_(m+1)-u_(n)`
Replace n by `k -2`
Then , `u_(k+1)=3u_(k)-(3)/(2)u_(k-1)-u_(k-2)`
`=(1)/(3)[3.2^k+3((1+sqrt(3))/(2))^k+3((1-sqrt(3))/(2))^k]`
`+(1)/(3)[-(3)/(2).2^(k-1)-(3)/(2)((1+sqrt(3))/(2))^(k-1)-(3)/(2)((1-sqrt(3))/(2))^(k-1)]`
`+(1)/(3)[-2^(k-2)-((1+sqrt(3))/(2))^(k-2)-((1-sqrt(3))/(2))^(k-2)]`
`=(1)/(3)[3.2^k .3.2^(k-2)-2^(k-2)+((1+sqrt(3))/(2))^k-(3)/(2)((1+sqrt(3))/(2))^(k-1)-((1+sqrt(3))/(2))^(k-2)+3((1-sqrt(3))/(2))^k-(3)/(2)((1-sqrt(3))/(2))^(k-1)-((1-sqrt(3))/(2))^(k-2)]`
`=(1)/(2)[2^(k-2)(3.4-3-1)+((1+sqrt(3))/(2))^(k-2) [3((1+sqrt(3))/(2))^2-(3)/(2)((1+sqrt(3))/(2))-1]+((1-sqrt(3))/(2))^(k-2)[3((1-sqrt(3))/(2))^2-(3)/(2)((1-sqrt(3))/(2))-1]]`
`=(1)/(3)[2^(k-2).8+((1+sqrt(3))/(2))^(k-2)[(3(1+sqrt(3))^2-3(1+sqrt(3))-4)/(4)]+((1-sqrt(3))/(2))^(k-2)[(3(1-sqrt(3))^2-3(1-sqrt(3))-4)/(4)]]`
`=(1)/(3)[2^(k+1)+((1+sqrt(3))/(2))^(k-2)[(10+6sqrt(3))/(8)]+((1-sqrt(3))/(2))^(k-2)[(10-6sqrt(3))/(8)]]`
`=(1)/(3) [2^(k+1)+((1+sqrt(3))/(2))^(k-2)((1+sqrt(3))/(2))^3+((1-sqrt(3))/(2))^(k-2)((1-sqrt(3))/(2))^3]`
`=(1)/(3)[2^(k+1)+((1+sqrt(3))/(2))^(k+1)+((1-sqrt(3))/(2))^(k+1)]`
This shows that the result is true for `n=k+1`. Hence , by the principle of mathematical induction the result is true for all `n in N`.