Asked by kai
find the sum of 1+4+16+...+256
Answers
Answered by
mathhelper
looks like a geometric series with
a = 1, r = 4
We don't know how many terms to add, so
a r^(n-1) = term(n)
1(4)^(n-1) = 256
4^(n-1) = 4^4
n-1 = 4, so n = 5
sum(n) = a(r^n - 1)/(r-1)
= 1(4^5 - 1)/(4-1) = 341
Or
since there are only 5 terms, we could just fill in the missing term, then add
1+4+16+64+256
= 341
a = 1, r = 4
We don't know how many terms to add, so
a r^(n-1) = term(n)
1(4)^(n-1) = 256
4^(n-1) = 4^4
n-1 = 4, so n = 5
sum(n) = a(r^n - 1)/(r-1)
= 1(4^5 - 1)/(4-1) = 341
Or
since there are only 5 terms, we could just fill in the missing term, then add
1+4+16+64+256
= 341
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