Find a simple expression for the sum of (k^2) * nCk from k=1 to n and prove your answer.?
960925
The answer you gave cannot possibly be right. Let n=1. Then 2^n * (n+2)C2 = 2^1 * 3C2 = 6, but [k=1, n]∑k²*nCk = 1² * 1C1 = 1, and 6≠1.\n\nI'll post a proof of the correct answer just as soon as I figure out what it is.\n\nEdit: after much rumination, and many used pieces of scratch paper, I have found a solution. \n\n[k=1, n]∑(k²nCk) = n(n+1)2^(n-2)\n\nIn proving this, I will assume that you are already familiar with the fact that (n+1)Ck = nCk + nC(k-1) and that [k=0, n]∑nCk = 2^n. If not, a good proof of the first is provided in the wikipedia article: http://en.wikipedia.org/wiki/Pascal%27s_rule . The second follows from the fact that this is the sum of the numbers on the nth row of my triangle, and each row has a sum twice that of the row before it (think about how each row is constructed).\n\nOur strategy in this proof will be to find a recurrence relation for each sum, and then solve this relation to obtain a general formula. First we will need to prove a lemma:\n\n[k=0, n]∑knCk = n2^(n-1)\n\nProof:\n\nFirst note that [k=0, 0]∑k0Ck = 0*0C0 = 0. Now we find a recurrence relation as follows:\n\n[k=0, n+1]∑k(n+1)Ck\nExtract the (n+1)th term:\nn+1 + [k=0, n]∑k(n+1)Ck\nExpand the binomial coefficient:\nn+1 + [k=0, n]∑k(nCk + nC(k-1))\nDistribute:\nn+1 + [k=0, n]∑knCk + [k=0, n]∑knC(k-1)\nExtract the 0th term from the right-hand sum (this term is 0):\nn+1 + [k=0, n]∑knCk + [k=1, n]∑knC(k-1)\nChange the indices of summation:\nn+1 + [k=0, n]∑knCk + [k=0, n-1]∑(k+1)nCk\nSince the term we extracted at the beginning is the nth term of the sum on the right, we can merge it into the sum:\n[k=0, n]∑knCk + [k=0, n]∑(k+1)nCk\nDistribute the k+1 term:\n[k=0, n]∑knCk + [k=0, n]∑knCk + [k=0, n]∑nCk\nSimplify:\n2[k=0, n]∑knCk + [k=0, n]∑nCk\nAnd finally:\n2[k=0, n]∑knCk + 2^n\n\nSo we are looking for an f that satisfies the recurrence relation:\n\nf(0)=0\nf(n+1) = 2f(n) + 2^n\n\nApplying this to itself:\n\nf(n+1) = 2(2f(n-1) + 2^(n-1)) + 2^n = 4f(n-1) + 2*2^n\nf(n+1) = 4(2f(n-2) + 2^(n-2)) + 2*2^n = 8f(n-2) + 3*2^n\nExtrapolating this pattern further:\nf(n+1) = 2^(n+1)f(0) + (n+1)2^n\nBut f(0)=0, so this is simply:\nf(n+1) = (n+1)2^n or f(n) = n2^(n-1)\n\nTo ensure our extrapolation is correct, we shall prove it rigorously by induction. First we note that f(0)=0, as it should. To show that n2^(n-1) satisfies f(n+1) = 2f(n) + 2^n in general, we assume it works for some n. Then:\n\nf(n+1) = 2f(n) + 2^n\nf(n+1) = 2n2^(n-1) + 2^n\nf(n+1) = n2^n + 2^n\nf(n+1) = (n+1)2^n\n\nWhich is what we get by plugging n+1 into the formula. So it works for all n.\n\nNow having proved this lemma, we proceed to [k=0, n]∑k²nCk (yes, your original formula had [k=1, n]∑k²nCk, however since the k=0 term is zero, including it does not change the sum). We'll use the function g to avoid confusing it with the previous function. Of course g(0)=0. Now we need our recurrence relation:\n\n[k=0, n+1]∑k²(n+1)Ck\nExtract the (n+1)th term:\n(n+1)² + [k=0, n]∑k²(n+1)Ck\nExpand the binomial coefficient and distribute:\n(n+1)² + [k=0, n]∑k²(nCk + nC(k-1))\n(n+1)² + [k=0, n]∑k²nCk + [k=1, n]∑k²nC(k-1)\nChange the indices of summation:\n(n+1)² + [k=0, n]∑k²nCk + [k=0, n-1]∑(k+1)²nCk\n(n+1)² is the nth term of the sum on the right, so we can merge them:\n[k=0, n]∑k²nCk + [k=0, n]∑(k+1)²nCk\nExpand the binomial:\n[k=0, n]∑k²nCk + [k=0, n]∑(k²+2k+1)nCk\nAnd distribute:\n[k=0, n]∑k²nCk + [k=0, n]∑k²nCk + 2[k=0, n]∑knCk + [k=0, n]∑nCk\nEmploying the formulas for [k=0, n]∑nCk and [k=0, n]∑knCk (the latter of which was proved as our lemma):\n2[k=0, n]∑k²nCk + 2n2^(n-1) + 2^n\nAnd simplifying further:\n2[k=0, n]∑k²nCk + (n+1)2^n\n\nSo g(n+1)=2g(n) + (n+1)2^n. Applying this to itself:\n\ng(n+1) = 2(2g(n-1) + n2^(n-1)) + (n+1)2^n = 4g(n-1) + ((n+1)+n)2^n\ng(n+1)=4(2g(n-2) + (n-1)2^(n-2)) + ((n+1)+n)2^n = 8g(n-2) + ((n+1) + n + (n-1))2^n\nExtrapolating this pattern further:
552266
that is like a^2-b^2=(a-b)*(a+b), so:\n9m^2-kn^2 must be equal to (3m+7n)*(3m-7n)=(9m^2-49n^2)\n<=> 9m^2-kn^2=9m^2-49n^2\n<=> 9m^2-9m^2+49n^2=kn^2 (we can do this because 9m^2>0)\n<=> 49n^2=kn^2 (also, because n^2>0, it can be divided with itself at both sides so we result in:)\n<=> 49=k
783793
The summation formula you used before,\n\n4n(n+1)/2 - 5n,\n\nsums 4x-5 from 1 to 33, giving you 2079.\n\nYou want to sum from 4 to 33, so you need to subtract off the terms corresponding to 1,2,3.  That is, you want\n\n4n(n+1)/2 - 5n - [ 4k(k+1)/2 - 5k ]\n\nwhere n=33, and k=3.
480863
There's something missing in that hint, there must still be an 'n' in there somewhere.\n\nTo work out the general formula:\n\nThe number of 2*2 matrices is (n-1)(m-1). The number of 2*3 is (n-1)(m-2). The number of 2*4 is (n-1)(m-3)... and so on until you reach (n-1)(1) = (n-1)(m-(m-1))\n\nSo the total number of 2 * j matrices is\nSigma(j = 1) { (n-1)(m-j) }\n\nThen the number of 3 * j matrices is\nSigma(j = 1) { (n-2)(m-j) }\n\nAnd so on, up to\nSigma(j = 1) { (1)(m-j) }\n\nThe sum of all these gives the number of i * j matrices as:\n\nSigma(i=1 to n-1)Sigma(j=1 to m-1) {(n-i)(m-j)}
810638
b=a+1\nc=a+2\nd=a+3\ne=a+4\nf=a+5\ng=a+6\nh=a+7\na^2+d^2+f^2+g^2\n=a^2+(a+3)^2+(a+5)^2+(a+6)^2\n=4a^2+28a+70\nb^2+c^2+e^2+h^2\n=(a+1)^2+(a+2)^2+(a+4)^2+(a+7)^2\n=4a^2+28a+70\nhence proved
775420
you can write \nP=d product((x-a_i)^k_i)) poduct((x-b_j)^2 + c_j^2)\n Then I give you 3 hints:\n1) d must be positive\n2) all k_i are even numbers\n3) use the formula (a^2+b^2)(c^2+d^2) =\n= (ac-bd)^2 + (ad+bc)^2
945958
(2n-1) * (2n-2) * (2n-3) * (2n-4)... * 3 * 2 * 1\n\nJust like any other factorial.
1139003
(n/(n - 3)) + (5/(2n - 6))\n(n/(n - 3)) + (5/(2(n - 3)))\nMultiply everything by 2(n - 3)\n(2n + 5)/(2n - 6)
748360
Regroup and add.  You got the 5n, so you should be able to get the other part!\n\nn + n - 1 + n - 2 + n + 7 + n - 11 = \nn + n + n + n + n - 1 - 2 + 7 - 11 =\n5n - 7\n\nNotice that -1 -2 - 11 = -14, add 7 to it and you get -7\n\nNRT
1032174
Observe:\nsum= 4+12+24+40+60+.....+t(n-1)+t(n)\nsum=       4+12+24+40+.....+t(n-2)+t(n-1)+t(n)\nSubtracting,\nt(n)=4+8+12+16+20+.........upto n terms\n      = n/2[2*4+(n-1)*4]\n      = n/2[4+4n]=n[2+2n]
1054159
First, get rid of the brackets (multiply the terms inside the brackets by -1):\n\n7-2x^2 +2x -2 -2x -5 +2x^2\n\nCollect like terms:\n\n2x^2 -2x^2 +2x -2x +7-2-5 = 0\n\nIt's easy to see the sum is zero.
5300
(n+k-1) over (k) = (n+k-1) / (k)\n\n=> (n+k-1) / (k) = n/k +1 -1/k
662979
If you expand (x+y)^n in terms of Binomial Theorem, you will see that\n\n(x+y)^n = C(n,0)*(x^n) + C(n,1)*(x^n-1)*y + ... + C(n,n)*(y^n)\n\nThe coefficients of this expansion are the C's which represent Combinations of values. Hence, sum of the co-efficients equals:\n\nC(n,0)+C(n,1)+...C(n,n) = 4096.\n\nUsing an simple mathematical proof, we can show that the sum of the above combinations equals 2^n. Thus,\n\n2^n = 4096\n\nSolving for n, we get:\n\nn = 12\n\n\nNow, to find the greatest cofficient of expansion, we look into all the coefficients together:\n\nC(12,0) = 1\nC(12,1) = 12\nC(12,2) = 66\n...\n...\nC(12,10) = 66\nC(12,11) = 12\nC(12,12) = 1\n\nYou will find a symmetry when you calculate the coefficients & you can easily judge that the maximum value lies exactly in the middle of the list, i.e. C(12,6)\n\nThus the greatest cofficient in the expansion is:\n\nC(12,6) = (12*11*10*9*8*7)/(1*2*3*4*5*6) = 924\n\nThus the greatest coefficient of expansio for your equation is 924 - which turns out to be the coefficient of (x^6)*(y^6)\n\n\n\n\n\nP.S. Oops... I feel kinda ashamed that I said 'advanced mathematical proof - it is actually a very simple proof as what 'geek24' said.
357472
Use the formula S(n) = k x (1- r^n+1)/(1-r) where k=6, r=3 and n = 17....the answer is 1,162,261,464
522619
1 1 1 2\n1 1 2 1\n1 2 1 1\n2 1 1 1\n\nNo other case is possible, So n(A)=4\n\n\n1 1 1 3\n1 1 3 1\n1 3 1 1\n3 1 1 1\n\n1 1 2 2\n1 2 1 2\n1 2 2 1\n\n2 1 1 2\n2 1 2 1\n2 2 1 1\n\nNo other possibility\nn(B)=10
859057
2+(n-1)2+1/2(n-1)(n-2)1\n= 2 + 2n - 2 + (1/2)(n^2 - 2n - n + 2)\n= 2n + (1/2)(n^2 - 3n + 2)\n= 2n + (1/2)n^2 - (1/2)(3n) + (1/2)2\n= 2n + (1/2)n^2 - (3/2)n + 1\n= (1/2)n^2 + (1/2)n + 1
694740
Is this for the sum of the first n cubes formula? I can tell you what your problem is right now: that two should be _inside_ the brackets. Otherwise, this formula doesn't work.\n\nNow, using the correct formula:\n(k(k+1)/2)² + (k+1)³\nWe start by actually multiplying out the stuff in the parentheses:\n((k²+k)/2)² + (k+1)³\n(k^4 + 2k³ + k²)/4 + (k³ + 3k² + 3k + 1)\nNow we give these fractions a common denominator:\n(k^4 + 2k³ + k²)/4 + (4k³ + 12k² + 12k + 4)/4\nAnd add:\n(k^4 + 6k³ + 13k² + 12k + 4)/4\nFactoring the polynomial:\n(k+1)²(k+2)²/4\n((k+1)(k+2)/2)²\nThus completing the proof.