what, you don't recognize
1,3,5,7,9,...
1,4,9,16,25,...
??
∑ (2k-1)/k^2
Express the um using sigma notation:
1 +3/4 +5/9+ 7/16+ 9/25
5 answers
maybe
a(1) = 1
a(n+1) = (a(n)+2)/n^2
sum = sigma from n=1 to n = 5 of a(n)
a(1) = 1
a(n+1) = (a(n)+2)/n^2
sum = sigma from n=1 to n = 5 of a(n)
oobleck gave better answer.
Thank you very much :)
What oobleck said, but with the upper and lower limits will give:
n=5
∑ (2k-1)/k^2
n=1
n=5
∑ (2k-1)/k^2
n=1
1 answer