SOLVING SIGNALS & SYSTEM PROBLEM

[rramya]

Please help:


(k=infinity)
Summation (1/3)^k (1/4)^(n-k) u[ n-k+3] is equal to
(k=0 )


(1/4)^n (1/11) - 3 (1/4)^n + 3 (256) (1/3)^n n>=3


i didn't get the ans correctly.

I solved as:

since :u[ n-k+3] term is involved:

n-k+3 >= 0
k <= n+3
implying n>=-3

now the summation becomes:

(k=n+3)
Summation (1/3)^k (1/4)^(n-k)
(k=0 )


next step:
(k=n+3)
(1/4)^ Summation (1/3)^k (1/4) ^(-k)
(k=0 )

next step:
(k=n+3)
(1/4)^ Summation (1/3)^k (4) ^(k)
(k=0 )

next step:
(k=n+3)
(1/4)^ Summation (4/3)^k
(k=0 )

next step:
(k=infinity) (k=infinity)
(1/4)^ [ Summation (4/3)^k - Summation (4/3)^k ]
(k=0 ) (k=n+4 )

next step:
(k=infinity)
(1/4)^ [ [ 1/(1-(4/3)) ] - Summation (4/3)^k ]
(k=n+4 )

next step:
(k=infinity)
(1/4)^ [ -3 - Summation (4/3)^k ]
(k=n+4 )

next step: (k=infinity)
(-3)(1/4)^ - [ (1/4)^ Summation (4/3)^k ]
(k=n+4 )

next step:
let p= k-n-4 so when k=n+4 ; p=0 ; when k=infinity ; p=infinity

next step: (p=infinity)
(-3)(1/4)^ - [ (1/4)^ Summation (4/3)^(p+n+4) ]
(p= 0)

next step: (p=infinity)
(-3)(1/4)^ - [ (1/4)^ (4/3)^(n+4) Summation (4/3)^(p) ]
(p= 0)


next step: (p=infinity)
(-3)(1/4)^ - [ 4^4 (1/3)^(n+4) Summation (4/3)^(p) ]
(p= 0)

next step:
(-3)(1/4)^ - [ 4^4 (1/3)^(n+4) ( 1 / (1-(4/3)) ) ]

next step:
(-3)(1/4)^ - [ 4^4 (1/3)^(n+4) ( -3) ]

FINAl step: ANSWER , I GOT:
(-3)(1/4)^ + [ 4^4 (1/3)^(n+3) ]

WHICH IS NOT THE FINAL ANSWER....

CAN ANY BODY POINT OUT MY MISTAKE..........