How to calculate the sum?
829342
sin(((2n+1)pi)/2)=-1if n is odd\nsin(((2n+1)pi)/2)=1if n is even\nso u get sum as\n-1/3+1/3^2-1/3^3+...............\nwhich is a GP\n=(-1/3)/[1-(-1/3)]\n=(-1/3)/{1+1/3}\n=(-1/3)/[4/3}\n=-1/4
323842
OK, the idea with this sum is that the questions just gives you ANOTHER NAME for the number 8, and you have to IMAGINE the number 8 when you see this OTHER NAME.\n\nHere is the sum we see that the 'other name' is going to be 'n'. So whenever we see an 'n' we have to think '8'.\n\nSo let's put this into the sum:\n\n12-n = ??\n\nso....\n\n12-8 = ??\n\nfrom here it's just normal taking away:\n\n12-8 = 4
854763
Not entirely sure what you're asking, but try this:\n\n3000 + (-710)  \n\nThe above is a sum of two numbers, one positive and one negative.\n\nOf course, hopefully you know that 3000 + (-710) = 3000 - 710.\n\n~ ♥ ~
680780
You can use infinite series...\n0 = 0+0+0+...\nsubstitute 0 = (1-1)\nso \n0 = (1 - 1) + (1 - 1) + (1 - 1) + ...\n\nuse associatve law of addition so that\n0 = 1 + (-1 + 1) + (-1 + 1) + (-1 + 1) + ... \n\nthen cancel out leaving you with \n0=1\n\nerror: using associative law on on infinite sum that does not converge
255108
Say x=0.93\nthen the sum simplifies to\n(x^6)(x^ -8)= x^ -2 = 1/x^2\n= 1/(0.93)(0.93)\n= 1.1562030292519366400739969938721
806827
depends what your trying to sum from the polynomial... what are you trying to get, theres to many thing just to say sum of given polynomials
201475
Do the math or at least show us the factors to get the sum and then we can decide whether or not it's true.  You can't get a sum without addends right?
1051751
Use yor calculator....
1067729
The sum of (-1)^n from 0 to infinite...1-1+1-1+1-1....\nGrouping this sum like this first: (1-1)+(1-1)+(1-1)...=0+0+0+...=0\nOR Use the following grouping: 1+(-1+1)+(-1+1)+...=1+0+0+0...=1\n\nThis is a fun problem, and shows that when using summation, be careful.
510030
OK, found the error:\n\nThe sum you have defined is sum(i=1 to n)i(n-i+1)  <== Note "+1"\n\n1. Let n = 1.\nThen 1(1) = 1\nand (1/6)(1)(1+1)(1+2) = 1.\n\n2. Let this be true for n.\nThen sum(i=1 to n)i(n-i) = (1/6)(n)(n+1)(n+2)\n\nSo for sum(i=1 to n+1)(i(n+1-i+1))  <== Note "+1"\n= sum(i=1 to n+1)(i(n-i+1)) + sum(i=1 to n+1)(i)\n= sum(i=1 to n)(i(n-i+1) + 0 + sum(i=1 to n+1)(i)\n\n2a) = n(n+1)(n+2)/6 + (n+1)(n+2)/2\n= (n+1)(n+2)/6 [ n + 3 ]\n= (n+1)(n+2)(n+3) / 6  ==> proof since this is what you expect from the theorem.\n\nEdit: Missing the +1 in the original summation is the issue.
166491
Start with adding 7 + (9 + 12).  The sum is 28.  The equation now looks like this:\n\n(c + 9) + 12 = 28\n\nNext, subtract 12 from both sides of the equation:\n\n(c + 9) + 12 - 12 = 28 - 12\n\nThe equation now looks like this:\n\nc + 9 = 16\n\nSolve it from there.
939729
Just "pair up" the terms.\n\n1 + (5-3) + (9-7) + (13-11) + ... + (2001-1999)\n\n1 + 2 + 2 + 2 + .... + 2 = 1 + 2 * 500 = 1001
883420
=A1+B1 for cells \nand for columns you could do \n=SUM(A1:A10)\n\nrows could be \n=SUM(A1:f1)
555590
Just multiply:\n(X/3)(3/4 x) -(x/3)(3/5) + (3/4)(3/4x) - (3/4)(3/5)\n\nand simplify from there
596283
add them up like this \n4+2+(-3)+6+5+(-3)+(-2)+0+(-1)+(-4)+3\ndivide the result by 11 because that is how many numbers there are in your list.  That will be your answer!
414270
Add the two tops and add two bottoms.Then the sum is\n(Top)[1+0 , 3+0] and (Bottom)[2+0 , 6+0] = E\nE is any 2x2 matrix and O is the 2x2 zeromatrix.\nAlways is E+O = O+E = E.\nThis is also true for nxn matrices.
440005
Pseudocode, I'm not up on my Java syntax:\n\nint sum=0;\nfor(int i=0;i<array.size;i++)\n  sum+=array[i];