finding the pattern of nth term?
475680
0 = 0^3\n1 = 1^3\n8 = 2^3\nand so on..\n125 = 5^3\n216 = 6^3\n\nSo it's 216
712174
term n = 2*3^n-1\nSo term 1=2*3^0 = 2\nterm 5 = 2*3^4 = 162
496619
Another way.\n\n=RIGHT(A1,LEN(A1)-SEARCH("source",A1,1)+1)\n\nHover your cursor over the three dots to see it in full.
594357
You asked for the nth term, so here it is\n\nFigure: n\n\nSquares: n(n+1)/2\n\nDots:  (n^2 + 5n + 2)/2\n\neg. n = 5\n\nsquares = 5 x 6/2 = 15\n\ndots: (25 + 25 + 2)/2 = 52/2 = 26
911326
T(n) - T(n-1) = log(n).\nT(n-1) - T(n-2) = log(n-1)\nT(n-2) - T(n-3) = log(n-2)\nT(n-3) - T(n-4) = log(n-3)\n...\nT(2) - T(1) = log(2)\n\nThe sum of all of these equations is\nT(n) - T(1) = log(n) + log(n-1) + ... + log(2) = log(n!)\nIsn't it great how nearly all of the terms on the left side cancel and all of the terms on the right collapse into one simple term?\nT(n) = log(n!) + T(1)\nT(n) = log(n!) + C
786690
u can do it by\n(n/2)(a+l)\nwhere \nn is no. of terms\na is first term\nl is last term
889816
1/rt3 *1/rt2-1/rt3\n[1/rt3](1/rt2-1)\n[1/rt3](1-rt2/rt2)\nrationalising\n[(1-rt2)*rt6]/6
183721
Let's first derive a formula.\nLet\na(1) = first term\na(n) = nth term\nn = number of terms\nd = common difference\ns(n) = sum of the first n terms\n\nThink of a sequence with nth terms.\nNext, add the 1st and nth(or last) terms.\nAdd the 2nd and (n - 1)th(or 2nd to the last) terms.\nAdd the 3rd and (n - 2)th(or 3rd to the last) terms.\n\nNotice something here? They are all equal. Eventually, you will add the nth and 1st terms as the last.\nall of them is equal to n[a(1) + a(n)] Notice that you've added each term twice. That means n[a(1) + a(n)] is twice the sum of the sequence, and the sum is:\ns(n) =  n[a(1) + a(n)]/2\n\nSo, in your sequence,\na(1) = 5\nn = 35\nd = 4\n\nuse the basic arithmetic recursion formula:\na(n) = a(1) + (n - 1)d\na(35) = 5 + (35 - 1)(4)\na(35) = 5 + 34*4\na(35) = 5 + 136\na(35) = 141\n\nwe can now use our derived formula:\n\ns(n) = n[a(1) + a(n)]/2\na(35) = 35[5 + 141]/2\na(35) = 35[146]/2\na(35) = 35(73)\na(35) = 2555\n\nHope this helps^_^
917224
Simply use correct order of operations\n\nparentheses\nexponents\nmult/division (from left to right)\naddition/subtraction (from left to right)
706696
EDIT:  Left_Field is right, n = no of periods.  Corrections below\n\n\nIt is done by calculus.  Continuous compounding is\n\nA = lim(n->∞) P*(1+r/n)^nt\n\n= lim(n->∞) P*[(1+r/n)^n]^t\n\nlim(n->∞) (1+r/n)^n = e^r, e is the base of the natural logs\n\nA = P*[e^r]t = P*e^rt
775652
It is a set of numbers arranged in a triangle. The first number which is located in the apex of the triangle is one. The next series of numbers under it is one, and one again. The third row of numbers is one, two and one. Then, one, 3, 3, and one. If you continue this, it would look like   \n                                      1\n                                 1      1\n                             1       2      1\n                          1     3        3     1\n\nAnd if you will notice, the 1st term and last term is always one. The second term is the sum of the previous numbers of the previous row. If you continue the trend, then the 4th row is 1, 4, 6, 4, 1.\n\nSee also www. wikipedia.com\n\n                                      1\n                                   1   1\n                                 1   2   1\n                               1   3   3   1\n                            1   4   6   4   1\n                         1   5  10  10   5   1
924613
http://www.mathwords.com/\n\n\nexcellent site.
338819
Again, it's derived from the last 2 letters in these words:  firST, secoND, thiRD, fourTH\n\nThe function stndrdth when used with a number such as a date gives the correct -st -nd -rd -th ending to that number
763343
Heartsense is right.  I'm chiming in to say if you mean the product of 5/6 * 6/7 * 7/8 * 8/9 * ..., then that product tends to 0 since the denominator of the nth term cancels out with the numerator of the (n+1)st term, so as n approaches infinity, you get 5 over infinity (not really "over infinity" but you know what I mean) = 0.
975361
d=nt2/p\npd=2nt\np=2nt/d
572436
Arithmetic sequence:  1, 2, 3, 4, 5, 6, 7\nGeometric sequence:  1, 2, 4, 8, 16, 32, 64
968502
Domain\n(-∞, ∞)\n\nRange\n[-3, ∞)