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find the indicated sum not sure how to type the sigma sign, but here is the problem 6 on top of the sigma sign. under the sigma sign is i=3 and on the right side of the sigma sign (3i2) 
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Express the sum using summation notation: 1  1/2 + 1/3  1/4 + 1/5  ... (to 3n terms) If I'm doing this right so far, with k=1 (the number on bottom of the sigma), the equation after the sigma would be ((1)^(k+1)) / k I'd be able … 
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Express the sum using summation notation: 1  1/2 + 1/3  1/4 + 1/5  ... (to 3n terms) If I'm doing this right so far, with k=1 (the number on bottom of the sigma), the equation after the sigma would be ((1)^(k+1)) / k I'd be able … 
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Express the sum using summation notation: 1  1/2 + 1/3  1/4 + 1/5  ... (to 3n terms) If I'm doing this right so far, with k=1 (the number on bottom of the sigma), the equation after the sigma would be ((1)^(k+1)) / k I'd be able … 
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hi i startted the problem below by writing a sigma notation of sigma[(2^(n+1))/((n+1)!)] QUESTION: how do i find the sum of 2+(4/2!)+(8/3!)+(16/4!)+... was my attempt wrong, or how do i go from here? 
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hi i startted the problem below by writing a sigma notation of sigma[(2^(n+1))/((n+1)!)] QUESTION: how do i find the sum of 2+(4/2!)+(8/3!)+(16/4!)+... was my attempt wrong, or how do i go from here? 
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Please check: 1.Write in expanded form: this is a sigma notation 4 on top signma in middle k=1 on bottom to right of sigma is (k1)(k2) I solved: k=1 (11)(12)=0(1)=0 K=2 (21)(22) = 1(0)=0 k=3(31)(32)=2(1)=2 k=4 (41)(4)=3(2)=6 … 
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I just need help with this. Write the series in sigma notation. 1/4 + 1/2+ 3/4 + 1 + 5/4 + 3/2 5 A.) 1/4 sigma N n=1 6 B.) 1/4 sigma n n=1 5 C.)sigma n/n+3 n=1 6 D.)sigma n/n+3 n=1 
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I just need help with this. Write the series in sigma notation. 1/4 + 1/2+ 3/4 + 1 + 5/4 + 3/2 5 A.) 1/4 sigma N n=1 6 B.) 1/4 sigma n n=1 5 C.)sigma n/n+3 n=1 6 D.)sigma n/n+3 n=1