- #1
dekoi
Show that:
[tex]\sum_{k=0}^{N} a_k = \sum_{k=M}^{M+N} a_{k-M}[/tex]
I did this, my answer is:
[tex]\sum_{k=0}^{N} a_k = a_0 + a_1 + a_2 + ... + a_N[/tex]
[tex]\sum_{k=M}^{M+N} a_{k-M} = a_{M-M} + a_{M+1-M} + a_{M+2-M} + ... + a_{M+N-M} = a_0 + a_1 + a_2 + ... + a_N[/tex]
Now, I have to use this to prove that:
[tex]\sum_{k=M}^{N} 2^{-k} = 2(2^{-M} - 2^{-N})[/tex]
So, I tried expanding the sum:
[tex]\sum_{k=M}^{N} 2^{-k} = 2^{-M} + 2^{-(M+1)} + 2^{-(M+2)} + ... + 2^{-N}[/tex]
I factored out some 2's:
[tex]\sum_{k=M}^{N} 2^{-k} = 2^{-M}2^{0} + 2^{-M}2^{-1} + 2^{-M}2^{-2} + ... + 2^{-M}2^{-N} = 2^{-M}(2^{0} + 2^{-1} + 2^{-2} + ... + 2^{-N})[/tex]
Which I'm assuming is equal to some other sum:
[tex]\sum_{k=M}^{N} 2^{-k} = 2^{-M}(\sum_{i=0}^{N} 2^{-i})[/tex]However, I don't know how to continue this. Any suggestions?
[tex]\sum_{k=0}^{N} a_k = \sum_{k=M}^{M+N} a_{k-M}[/tex]
I did this, my answer is:
[tex]\sum_{k=0}^{N} a_k = a_0 + a_1 + a_2 + ... + a_N[/tex]
[tex]\sum_{k=M}^{M+N} a_{k-M} = a_{M-M} + a_{M+1-M} + a_{M+2-M} + ... + a_{M+N-M} = a_0 + a_1 + a_2 + ... + a_N[/tex]
Now, I have to use this to prove that:
[tex]\sum_{k=M}^{N} 2^{-k} = 2(2^{-M} - 2^{-N})[/tex]
So, I tried expanding the sum:
[tex]\sum_{k=M}^{N} 2^{-k} = 2^{-M} + 2^{-(M+1)} + 2^{-(M+2)} + ... + 2^{-N}[/tex]
I factored out some 2's:
[tex]\sum_{k=M}^{N} 2^{-k} = 2^{-M}2^{0} + 2^{-M}2^{-1} + 2^{-M}2^{-2} + ... + 2^{-M}2^{-N} = 2^{-M}(2^{0} + 2^{-1} + 2^{-2} + ... + 2^{-N})[/tex]
Which I'm assuming is equal to some other sum:
[tex]\sum_{k=M}^{N} 2^{-k} = 2^{-M}(\sum_{i=0}^{N} 2^{-i})[/tex]However, I don't know how to continue this. Any suggestions?