Find the first and last terms of the following sequence:?
815660
Easy!!!\nYou're all wrong! it's not an arithmetic sequence! you have just put logic into it and just add & minus and use your common sense!!\n\nBut the pattern is a pythagorean triple!! so the answers are:\n\n√5 and 3612!!\n\nWhy?\nsimply because terms 1,2,3 form a pythagorean triple!\n\nCheck!\n(√5)² + 2² = 3² is the equation true?\n5 + 4 = 9\n9=9 TRUE!!!\n\nso the answer is √5\nnext terms!\n3 4 and 5\nCheck:\n3² + 4² = 5²\n9+16=25\n25=25 TRUE!!\n\nNext!\n5, 12, 13\nCheck:\n5²+12²=13²\n25+144= 169\n169=169 TRUE!\n\n\nNext!\n13,84,85\n\n13²+84²=85²\n169 + 7056 = 7225\n7225=7225 TRUE!\n\nLAST!!\n85,3612,3613\n\n85² + 3612²= 3613²\n7225 +13046544 = 13053769\n\n13053769 = 13053769  TRUE!!!\n\nSO the answers are!!\n\n√5 and 3612!\n\nThat's all!! ^_^!!
769753
Without knowing the relationship between successive elements, I think the best you can do is this:\n[k=0, n-1]∑(x_(k+1) - x_k)²\n[k=0, n-1]∑(x_(k+1)² - 2x_k*x_(k+1) + x_k²)\n[k=0, n-1]∑x_(k+1)² - 2[k=0, n-1]∑(x_k*x_(k+1)) + [k=0, n-1]∑x_k²\n[k=1, n]∑x_k² - 2[k=0, n-1]∑(x_k*x_(k+1)) + [k=0, n-1]∑x_k²\n\nx_0 + x_n + 2[k=1, n-1]∑x_k² - 2[k=0, n-1]∑(x_k*x_(k+1))
168953
121    since(1+4)+(1+4*2)+(1+4*4)
1079282
If by a(i) and b(i) you mean "a times i" and "b times i", then:\n\nFor i = 0:\n\na*i / (b*i + 1) = 0 / 1 = 0\n\nSo you have:\n\nPo = 1 / (1 + 0 + (0 * (a / (b + 1))) + (0 * (a / (b + 1)) * (2a / (2b + 1))) + ...)\n\nPo = 1 / 1\n\nPo = 1\n\nIf by a(i) and b(i) you mean a and b are infinite sequences (then presumably a / b means "for all i, a(i) / b(1) < 1"), well then that is something else entirely.\n\nThis reminds me of the problem to expand and simplify the product of twenty-six binomial terms:\n\n(x - a) (x - b) (x - c) ... (x - z)
483473
there might be a - or a . in the sequence
504275
1. The equation for the first would be                      \n\nt(n)={2^(5-1)}*{3^(n-1)}   It is a A.P.\n\n2. It is neither
108578
the first term is A11=8/3^9\nAn = A1*q^(n-1)\nin our sequence A1=24 and q=3\ntherefore An=24*(1/3)^(n-1)\nlet's say 0.001 = 24*(1/3)^(n-1)\n0.001/24 = (1/3)^(n-1)\n1/24000 = ((1/3)^n)/(1/3)        /:1/3\n1/72000 = (1/3)^n\nn = (log(1/72000))/(log(1/3))\nn =  10.18\nwe want An to be less then 0.001 so we take a greater integer n, so n = 11 and A11 = 24*(1/3)^10 = 8/3^9
875476
(a + b - c)(a + b + c)\n((a+b)+c)((a+b)-c)\n(a+b)^2-c^2\na^2+2ab+b^2-c^2
500014
a. (x+9)(x-9)\nb.(t-2)(t+2)\nc. (b-12)(b+12)
8373
(([(4+4+4).(4+4)]+4)/4)*(4+4)\n\n(4+(4*4)).([(4+4+4).(4+4)]+4)
868097
-\naside from 1,\n2,3,4,6,12
187387
(n)(3n-1)(3n-2).......(n+1)
502428
#1. (x-5)(x+1)\n#2. (x-3)(x+2)
952696
n + (n - 5) + (n + 6) \n3n + 1
222592
1. p'q\n\n2. p'\n\n3. pq' + p'\n\nI think i got the correct ones !
187653
in geometric sequence \nUn = U1[r^(n-1)]  \nwhere,\nU1 = first term\nr = common ratio\nn = number of terms\n\nthen we can say that,\nU4 = U1[r^(4-1)]\n24 = U1[r^3]    -> let this be equation (1)\nand\nU7 = U1[r^(7-1)]\n192 = U1[r^6]   -> let this be equation (2)\n\nif you divide equation (1) by equation (2), you'll get\n\n(24/192) = U1[r^3] / U1[r^6]\n(1/8) = 1/r^3\ntherefore \nr = 2\nif r=2 then you can get U1 by substituting r in equation (1)\n\n24 = U1[r^3]\nU1 = 24/8\nU1 = 3\n\ntherefore,\n\nUn = 3[2^(n-1)]
859715
1) 4-3x<_8+x\n4+8<_x+3x\n12<_4x\n3<_x or x in the interval [3,plus infinity)\n\n2) 3x-3>-x+1\n3x+x>1+3\n4x>4\nx>1 or x in the interval (1, plus infinity)\n\n3) 3(x-1)_<4x\n3x-3<_4x\n-3<_x or x in the interval [-3, plus infinity)\n\n4)1/2x<x+1\n-1<x/2\n-2<x or x in the interval (-2, plus infinity)\n\n5)-1/2x>1-x\nx/2>1\nx>2 or x in the interval (2, plus infinity)