using the formula for the sum of the first n terms of a geometic series what is the sum of the first 10 terms four decimals? what is the sum for the first 12 terms?
213741
The common ratio is (1/2)/1 = 1/2\nFormula for sum of terms is a(1-r^n)/(1-r) , where a is the first term, r- common ratio and n is the number of terms.\n\nsum of first ten terms is 1(1-(1/2)^10)/(1-1/2) = 1.9980\n\nsum of first twelve terms is (1- (1/2)^12)/(1-1/2) = 1.9995\n\nif n tends to infinity the sum becomes a/1-r  , in this case 1/(1/2) = 2
1079
What you are describing is an infinite series using a decreasing geometric progression. The sum of such a series can be described by this proof:\n\nLet T = total distance traveled (in spaces)\nThen T = 1 + (1/2) + (1/4) + (1/8) + (1/16) ...\nAlso T - 1 = (1/2) + (1/4) + (1/8) + (1/16) ...\nAnd if we multiply both sides of the equation by 2:\n2*(T - 1) = 2*((1/2) + (1/4) + (1/8) + (1/16) ...)\nMultiplying through:\n2T - 2 = 1 + (1/2) + (1/4) + (1/8) + (1/16) ...\nBut the Right side is the same as T!\n2T - 2 = T\nSimplifying:\nT = 2\nSo the total distance traveled is 2 spaces.
392256
For the first one: yes, you can multiply the expression term by term. If f(x) is a function given by [n=0, ∞]∑(a_n x^n), and g(x) is any function, then f(x) * g(x) = [n=0, ∞]∑(g(x) a_n x^n). g(x) is just a constant in terms of n, so distributing it across the terms of the series will not change its convergence properties.\n\nFor the second question: you seem to be describing a Laurent series. These series are described in more detail on the Wikipedia:\n\nhttp://en.wikipedia.org/wiki/Laurent_series
509821
OK, let's see.\n\nIt's an arithmetic sequence, so the difference between consecutive terms is a constant.  The first term is 1, so we can express each term as:\n\nx(i) = 1 + k*(i-1), i = 1, 2, 3, ....\n\nNext the problem says that the 2nd, 10th, and 34th terms are the first three terms in a geometric sequence.  That means that the ratio of the 10th term to the 2nd term must be the same as the ratio of the 34th term to the 10th, because when you divide any two terms of a geometric sequence, you get the same constant.\n\ni(2) = 1 + 1*k\ni(10) = 1 + 9*k\ni(34) = 1 + 33*k\n\nSo we said i(10)/i(2) must equal i(34)/i(10):\n\n1+9k    1+33k\n-------- = --------\n1+1k      1+9k\n\nCross multiplying: (1+9k)^2 = (1+1k)(1+33k).\n\nThen 1 + 18k + 81k^2 = 1 + 34k + 33k^2.\n\nPutting it all on the left side, 48k^2 - 16k = 0, or 16k(3k - 1) = 0.\n\nSo either k = 0 (not interesting) or k = 1/3.\n\nTherefore the arithmetic sequence must be:\n\n1, 1 1/3, 1 2/3, 2, 2 1/3, 2 2/3, 3, 3 1/3, ....\n\nThe second term is 1 1/3.  The 10th term is 4 (1 + 3 * 1/3).  The 34th term is 12 (1 + 33 * 1/3).  So the geometric progression would be:\n\n4/3, 4, 12, 48, 192, .....
60091
Exponentiation is usually taken to bind more strongly than addition, so 10^n+1 is usually interpreted as (10^n)+1.  You intended 10^(n+1).\n\nLet Rep(7, 10, n) be the "repunit" the integer the base 10 representation of which is n seven's.  You wish to prove by Mathematical Induction that the statement P(n) defined as\n\ndefine: P(n) states Sum{i = 1 to n}{Rep(7, 10, i)} = (7/81)*[10^(n+1) - (9*n) - 10]\n\nholds for all positive integers n = 1, 2, 3, ....\n\n\nFirst prove the lemma that Rep(7, 10, i) = "i seven's in base ten" = Sum{k = 0 to i-1}{7 * (10^k)} = 7 * ((10^i - 1) / 9) for positive integer i = 1, 2, 3, ... either by the formula for the sum of a geometric progression or by Mathematical Induction. :-)\n\n\nNow consider the statement P(1).  The left hand side of the equality is Sum{i = 1 to 1}{Rep(7, 10, i)} = Rep(7, 10, i) = 7.  The right hand side of the equality is (7/81)*[10^(1+1) - (9*1) - 10] = (7/81)*(100 - 9 - 10) = (7/81)*81 = 7.  The left hand side equals the right hand side so P(1) is true.\n\nNow let m be a positive integer m >= 1 and assume the statement P(m) is true with the goal of proving that the P(m+1) must then also be true.  For the statement P(m+1) the left hand side of the equality is\n\nSum{i = 1 to m+1}{Rep(7, 10, i)} = Sum{i = 1 to m}{Rep(7, 10, i)}  +  Rep(7, 10, m+1)\n= Sum{i = 1 to m}{Rep(7, 10, i)}  +  7 * ((10^(m+1) - 1) / 9)      # by the lemma\n= (7/81)*[10^(m+1) - (9*m) - 10]  +  7 * ((10^(m+1) - 1) / 9)      # by the assumption of P(m)\n= (7/81)*[10^(m+1) - (9*m) - 10]  +  7 * (9 * (10^(m+1) - 1) / 81) # multiply both numerator and denominator of right addend by 9\n= (7/81)*[ (10^(m+1) - (9*m) - 10)  +  (9 * (10^(m+1) - 1)) ]      # "undistribute" common factor of 7/81\n= (7/81)*[ (10^(m+1) - (9*m) - 10)  +  (9 * 10^(m+1) - 9) ]        # distribute multiplication by 9\n= (7/81)*[ ( 10^(m+1) + 9*10^(m+1) ) - 9*m - 9 - 10 ]              # addition is commutative and associative\n= (7/81)*[ 10 * 10^(m+1) - 9*(m+1) - 10 ]                          # more distributuve law\n= (7/81)*[ 10^((m+1)+1) - 9*(m+1) - 10 ]                           # definition of exponentiation\n= right hand side of the equality of P(m+1)\n\nThus both P(1) is true and "for all integers m >= 1, P(m) implies P(m+1)" is true.  You can knock over the first domino and each domino when it falls will knock over the next.  So by Mathematical Induction "for all integers n >= 1, P(n)" is true and all of the dominoes eventually fall.
571655
9x4 + 82x6 - 24x12 + 87\n\n36 + 41x12 - 24x12 + 87\n\n36 + 17x12 + 87\n\n36 + 120 + 84 + 87\n\n156 + 84 + 87\n\n240 + 87\n\n327
481707
You're answer is perfectly correct.  You do want to use root test, and you applied it properly.\n\nYou could also check your work by sorta "thinking" it out.\n\nthe first term (if you start with n=1) is 1, then 1/4 then 1/27 then 1/256 and so on, so as you can see, if you sum the first 3 terms we get a number close to 1.3, if we sum the first 4 numbers its still about 1.3... doesn't matter how many terms it is, i think it ends up around 1.3 or 1.33333 or something, there is a way to figure out what the sum converges to.
290936
Dear amit.wizkid, thou little knoweth what an abysmal error thou maketh.\n\ntut_einstein, you are better off noting that the sum of the first 'n' terms of an AP, whose first term is 'a' and the difference between terms is 'd' is :\n\nS = (n/2)[2a + (n-1)d]\n\nYou state that \nS1/S2 = (n+1)/(n+3)\n\nwhich is \n\n[2a1 + (n-1)d1] :[2a2 + (n-1)d2] = (n+1) : (n+3) ..... (I)\n\nRemember that what you desire to know is\n(a1 + 5d1) : (a2 + 5d2) \n\nPut n = 11 in (I)\n\n[2a1 + 10 d1] :[2a2 + 10 d2] = 12:14\n\nwhich gives\n\n[a1 + 5d1] : [a2 + 5d2] = 6:7\n\nwhich is what you desire. \n\nShould suffice.
531430
1) you need to know what product and sum means. sum is addition and product is multiplication. Difference is subtraction.\n(10*6)-(7+8)\n\n2). Square has 4 equal sides. If the perimeter is 36 divide that by 4 and there is your answer.\n\n3). You need to know the hundreths place. tenths (.1) hundreths (.01)\n\n4). 1/4? that is 25%\n\n5). standard form is written like 1.000X10^x
178868
The formula for finding the nth term of an arithmetic sequence is\n\nan = a1 + (n - 1)d\n\nYou need the spacing, d:\n\n19 = 5 + (8 - 1)d\n14 = 7d\nd = 2\n\nSo now you can find the 100th term:\n\na(100) = 5 + (100 - 1)2\n= 5 + (99)2\n= 203
478226
I understand your confusion, and suggest that you focus on how the two problems are grouped.\n\nIn mathematics, it's customary to group added terms together using parentheses.  Therefore, the first sum is merely 6.\n\nOn to the next problem:  With nine 1's being added, the sum is 9.\n\nNever would you string these together to form a two-digit answer.\n\n\nHowever, as the first Answerer suggests, perhaps your problem was a single multiplication problem.  In that case (and assuming the problem was written all on one line), you would work it out as he/she did:\n\n (6)(9) = 6 * 9 = 54\n\nHope this helps!
18132
Only one side is correct, the other is FALSE (for both sum and product)!\n\nYou can have n irrational numbers (n>1) with a rational sum:\nx1=x2=...=x(n-1)=sqrt(2), xn=-(n-1)*sqrt(2)\nx1+x2+...+xn=0\n\nYou can also have n irrational numbers (n>1) with a rational product:\nx1=x2=...=xn=2^(1/n)\nx1*x2*...*xn=2\n\nThe correct side (when all numbers are rational, their sum and product are rational) is trivial, you can prove it through induction.
480054
Let F(x) = Sum n=0toInfty [anx^n] be the generating function of the sequence (an). Written in terms of F, the recurrence relation and the initial condition a0 =1 mean that, for every x (within the radius of convergence) \n\nF(x) = 1 + 8xF(x) + Sum n=0toInfty [10^n x^(n+1)].\n\n(To see this, just use the RR to substitute for an in terms of a(n-1) in the series that defines F; you get the same series but times 8x, plus the other bits.)\nThis gives an equation for F(x) which solves to \n\nF(x) = (1-9x)/[(1-8x)(1-10x)]\n\nExpand this into partial fractions \n\nF(x) = (1/2)[1/(1-8x) + 1/(1-10x)]\n\nand rewrite these as sums of geometric series to get \n\nF(x) = Sum n=0toInfty [(1/2)(10^n + 8^n) x^n]\n\nfor every x, which means that \n\nan = (1/2)(10^n + 8^n)\n\nfor every n.
406537
u can do it by\n(n/2)(a+l)\nwhere \nn is no. of terms\na is first term\nl is last term
369888
Even though somebody told you the answer, I'm going to explain how to do it so you know on other problems. \n\n\n4(3y+2m+1)=\n\nYou have to take the number outside of the parenthesis, and multiply it to all the other numbers.\n\nFirst you do 4 x 3y, because 3y is the first term. \nThat comes out to 12y. \n\nThe next one is 2m, so 4 x 2m= 8m\n\nThe last term is 1, so 4 x 1=4.\n\nAdd up all the numbers and the answer comes out to 12y+8m+3\n\nApply the same method to the 2nd problem\nHope this helps!
438972
a) not necessarily...\nb)if the bottom contains n cells, the number of stories is also n...\nand number of cards is 2n+3(n-1)+3(n-2)....+3(n-(n-1))....\nor 2n + 3+ 3(2) + 3(3)...3(n-1)  (writing from back...)\nthe second is a arithmetic series,with a =3 and d=3\nthe nth term is (3 +(n-1)*3)\nsince the formula for the sum upto n-1th term is \ns = [(n-1)*(2a+ (n-2)d)]/2 = [(n-1)*(6+(n-2)*3]/2 \n  = [3n^2 - 3n ]/2\nsummation of 2n and this is ...\n[4n + 3n^2 -3n]/2  =[ 3n^2 + n]/2 \n but you know total cards is 104.\nor    [3n^2 + n]/2  = 104\nor    3n^2 + n   = 208\n  3n^2 + n-208 = 0\nsolve this quadratic in n to get number of stories...\nwe get n =  8.16  or approx 8\nso we get number of cards used is\nfor bottom  = 2*n = 16\n1st storie  =  3* (n-1) = 21\n2nd storie  = 18\n3rd storie  = 15\n4th = 12\n5th = 9\n6th =6\n7th = 3\nadding  3+6+9+12+15+18+21+16, we get, 100...\nthus 4 cards go to waste for making the pyramid which has 8 stories (including bottom ) and uses 100 cards.\nyaaaaaaaaay  :D
117943
This is a geometric sequence with a common ratio of (-1/4).\n\nLet:\n\nr = common ratio\na1 = 1st term = 80\na2 = 2nd term = -20\na3 = 3rd term = 5\na4 = 4th term = -5/4\na5 = 5th term \na6 = 6th term\na7 = 7th term\n.\n.\n.\na10 = 10th term\n\nSolution:\n\nr = a2/a1 = a3/a2 = a4/a3 = -20/80 = -1/4\n\na1 = 80 Given\na2 = (a1)(r)= 80r = -20 Given\na3 = (a1)(r)(r) = 80r^2 = 5 Given\na4 = (a1)(r)(r)(r) = 80r^3 = -5/4 Given\na5 = (a1)(r)^(5-1) = 80r^4 = 80(-1/4)^4 = 5/16 ANSWER\na6 = (a1)(r)^(6-1) = 80r^5 = 80(-1/4)^5 = -5/64 ANSWER\na7 = (a1)(r)^(7-1) = 80r^6 = 80(-1/4)^6 = 5/256 ANSWER\n.\n.\na10 = (a1)(r)^(10-1) = 80r^9 = 80(-1/4)^9 = -5/16384 ANSWER