Precalculus
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posted by Steve
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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
asked by Please Help on June 12, 2010 
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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
asked by Anonymous on June 11, 2010 
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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
asked by Anonymous on June 10, 2010 
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It's in sigma notation, so I'm not sure how to type it. I'll just use E lol. Find each sum. 6 E (n+1)/(n+2) n=1 Is this a geometric or arithmetic series? That is pretty much all I really need to know. I appreciate all the help you
asked by matt on June 16, 2009 
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The roots of the eqn, x^4 + px^3 + qx^2 + rx + s = 0 where p, q, r, s are constants and s does not equal to 0, are a, b, c, d. (i) a^2 + b^2 + c^2 + d^2 = p^2 2q (in terms of p & q) (ii) 1/a + 1/b + 1/c + 1/d = r/s (in terms of
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Q: What represents the sum of a geometric series with 8 terms whose first term is 3 and common ratio is 4. A: 65,535 Is my answer right? I wasn't sure when adding up the terms versus using Sigma notation.
asked by Bill on March 7, 2016 
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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? thank you!
asked by kiksy on March 17, 2011 
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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? thank you!
asked by kiksy on March 17, 2011 
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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
asked by Bethany on December 30, 2011 
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Write the sum (in sigma notation) 1 + 3/2 + 2 + 5/2 + 3 + 7/2 + 4 Sigma notation?
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