anonymous
  • anonymous
Summation Evaluate these summations. a) N(E)k=1 (k-k^3) b) N-1(E)k=1 (k^2/N^2) Note that (E) is the summation symbol. Not sure how to do these. Help would be appreciated
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
hey m not sure i understand..r the terms on the right the expression for the kth term??
anonymous
  • anonymous
do we have to evaluate the sum of the series whose term is k-k^3????clarify and i cn help
anonymous
  • anonymous
sorry it's all squashed together, didn't notice! there are two questions: a) N (E) k=1 (k-k^3) b) N-1 (E) k=1 (k^2/N^2) where N = natural numbers and (E) is the summation symbol The N and N-1 for the questions are on top of the summation symbol and the k =1 are below the symbol. The (k-k^3) and (k^2/N^2) are on the right sides of the summation symbol.

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anonymous
  • anonymous
u can use the equation tab to write it in mathematical format
anonymous
  • anonymous
ugghh it keeps squashing together - but a) and b) are separate questions
anonymous
  • anonymous
N(N+1)/2 + [N(N+1)/2]^2
anonymous
  • anonymous
thats a)
anonymous
  • anonymous
as for b) (N-1)(N)(2N-1)/6N ^2
anonymous
  • anonymous
\[\sum_{k=1}^{N}(k-k^3)\]
anonymous
  • anonymous
yes uzma that's the correct format for a)
anonymous
  • anonymous
and b?????
anonymous
  • anonymous
\[\sum_{k}^{N-1}(k^2/N^2)\]
anonymous
  • anonymous
i gave d answers
anonymous
  • anonymous
Yes that's the correct one for b) him1618, thanks for your help :) and uzma too :)
anonymous
  • anonymous
no prob
anonymous
  • anonymous
did u understand the solution?
anonymous
  • anonymous
Yes, thank you :)
anonymous
  • anonymous
welcome:)
anonymous
  • anonymous
wht tym is it there at ur place?
anonymous
  • anonymous
9:02pm
anonymous
  • anonymous
where r u?
anonymous
  • anonymous
new zealand
anonymous
  • anonymous
k...
anonymous
  • anonymous
N(E)K=1(K-K^3) (E)1/1-K^2=1/N N=1-K^2 K=1 N=0
anonymous
  • anonymous
what sort of followup is this kihaga???
anonymous
  • anonymous
Yes! that is Honorata kihaga

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