(n 1) + (n 5) + (n 9) + (n 13) + ... sum of binomial coefficients
#(n 1) + (n 5) + (n 9) + (n 13) + ... sum of binomial coefficients
frail swallow
icy crownBOT
flint seal
What are you asking?
thorny creek
(n 1) + (n 5) + (n 9) + (n 13) + ... sum of binomial coefficients
You may expand (1+i)^n-(1-i)^n+((1+1)^n-(1-1)^n)*i and divide by 4i.
When simplified you get an answer like 2^(n-2)+2^(n/2-1) * sin(pi * n/4).
You can remove sin If you take cases of n mod 4.
frail swallow
what does it mean (1-1)^n??
frail swallow
You may expand (1+i)^n-(1-i)^n+((1+1)^n-(1-1)^n)*i and divide by 4i.
(1-1)
thorny creek
expand it as binomial. though it equals 0 of course
frail swallow
I get ([2*(n 1)i - 2(n 3)i + 2(n 5)i - 2(n 7)*i +...] + (2^n)*i) /4i
thorny creek
just write binomial expansions in 4 lines:
(1+i)^n
-(1-i)^n
i*(1+1)^n
-i*(1-1)^n
and then add them together.
frail swallow
(n 3) + (n 6) + ... it seems to be in the way
thorny creek
(n 3) + (n 6) + ... it seems to be in the way
nope