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  1. ...1 maybe > 1). 2. Assume it to be true for a generic term n. 3. Show that if it works for n it...2/2]^2 = 1 hence true for n = 1 assume true for n: ∑ j^3 = [n(n+1)/2]^2 for i = 1 to n for n+1 ∑ j^3...

    1 Answers · Science & Mathematics · 11/10/2009

  2. general term is a[j] = (n+1-j)*j; s(n) =Σa[j], where j=1 until n; ♠check n=2...3+1)*(3+2)/6 =10; good! ♣ s(n+1) =Σ ((n+1) +1 –j)*j = Σ ((n+1 -j +1)*j {j=1 until n+1} = = Σ((n+1 –j)*j +j...

    2 Answers · Science & Mathematics · 04/09/2008

  3. ... that, if you have a matrix of the form A = [a_1...ca_i + db_i...a_n] where the a_j's (and ca_i + db_i) stand for the columns of the matrix A and c...

    4 Answers · Science & Mathematics · 11/08/2008

  4. 1) (1, j) is a member of S for each j from N ==> cardinality of S is at least as big as the cardinality of N...

    1 Answers · Science & Mathematics · 21/07/2010

  5. n(j(x)) = 5/((-6 + 20/(-2 + x)) (-2 + x)) = 5/(32 - 6 x)

    1 Answers · Science & Mathematics · 30/07/2015

  6. ...j ∈ Z, m=3j } and B = {n ∈ Z ; ∃ k ∈ Z, n=7k} then A n B = {m ∈ Z ; ∃ j ∈ Z, m = 21j } 2. P({4,5,6}) = { {4,5;6} , { 4,5 } , { 4,6 } , { 5,6...

    1 Answers · Science & Mathematics · 22/12/2013

  7. ...list of all a[k] ^ j will still be completely distinct from the list n[k] and therefore (a^j, n) = (a, n) = 1

    2 Answers · Science & Mathematics · 31/01/2012

  8. ...proposition. Suppose this is true for n = k. Given p | a_1a_2a_3...a_k a_(k+1) = (a_1a_2a_3...a_k) * (a...inductive hypothesis implies that p | a_j for some j = 1,2,...,k. I hope that helps...

    3 Answers · Science & Mathematics · 08/09/2009

  9. ∑(n on top, i=1 bottom) a_i = a_1 + a_2 + ... + a_n but ∑(n top, i=1 bottom) a_ j = a_ j + a_ j + ... + a_ j = n * a_ j, since j is different from the index variable i. I hope that helps!

    1 Answers · Science & Mathematics · 26/07/2009

  10. ...n) has a subsequence (n_j) such that sin(n_j) → -1, which implies a_n_j = n_j^sin(n_j) → 0. We see a_n has a subsequence that goes...

    1 Answers · Science & Mathematics · 02/10/2008

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