math sequence and series

Sequences and Series - A sequence is an ordered listing of numbers such as {1, 3, 5, 7, 9}. In mathematical terms, a sequence is a function whose domain is the natural number set {1, 2, 3, 4, 5, …}, or some subset of the..

Sequences and Series - A sequence is an ordered listing of numbers such as {1, 3, 5, 7, 9}. In mathematical terms, a sequence is a function whose domain is the natural number set {1, 2, 3, 4, 5, …}, or some subset of the natural numbers, and whose range is the set of real numbers or some subset thereof. Let f denote the sequence {1, 3, 5, 7, 9}. Then f (1) = 1, f (2) = 3, f (3) = 5, f (4) = 7, and f (5) = 9. Here the domain of f is {1, 2, 3, 4, 5} and the range is {1, 3, 5, 7, 9}...

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  Can anyone explain final part of this AS level maths question on sequences and series?Daniel Kleinman creates one of the most nuanced layered sequences in the series for the action adventure thriller GoldenEye (1995), starring Pierce Brosnan in his first outing as James Bond. With the franchise returning after a seven year hiatus and many questioning 007's purpose, the film addresses the issue head on by reflecting the new geopolitical structure of a post Cold War world, with the focus being on broken and fractured Communist Russian imagery and the dangers therein. This ...

Question : The sequence of real numbers a[1], a[2], a[3], ... is an arithmetic progression. Prove, by mathematical induction, that summation( n , r=1 , r a[r] ) = n(n+1)( a[1] +2 a[n] ) / 6 for all positive integers of n. But how do you know the sequence a[1], a[2], ...? Could you explain further? Especially the first few steps. Thanks!

Answer : This is for a^2? not a? I can prove the a^2 if you need. Base Case: n = 1 1 = [1(1 + 1)(2(1) + 1)]/6 = 1 Assume it is true for n Show n+1: 1 + 2 + ... + n + (n + 1) = [n(n + 1)(2n + 1)]/6 + (n + 1) 1 + 2 + ... + n + (n + 1) = [n(n + 1)(2n + 1) + 6(n + 1) ]/6 1 + 2 + ... + n + (n + 1) = (n + 1)[(n(2n + 1) + 6(n + 1)]/6 1 + 2 + ... + n + (n + 1) = (n + 1)[(2n + n + 6n + 6)]/6 1 + 2 + ... + n + (n + 1) = (n + 1)[2n + 7n + 6]/6 1 + 2 + ... + n + (n + 1) = [(n + 1)(n + 2)(2n + 3)]/6 Since it works for n+1, it works for all whole numbers n (forgot how to write that correctly). Hope that helps, Mackler

Question : 1.) Determine the value of x such that x+2, 2x+3, 4x-3 are consecutive terms of an arithmetic sequence. (There's a part b to this, which is for a geometric sequence...but I should know what to do once I realize what to do for arithmetic) 2.) Three numbers form an arithmetic sequence, the common difference being 11. If the first number is decreased by 6, the second number is decreased by 1, and the third number doubled, the resulting numbers form a geometric sequence. Determine the numbers..

Answer : (1) (2x + 3) - (x+2) = (4x - 3) - (2x + 3) (Why?) (2) Let the first number be x, then the next two are (x+11) and (x+22). First decreased by 6: (x-6) Second increased by 1: (x+12) Third doubled: (2x + 44) Then (x+12) / (x-6) = (2x+44) / (x+12) (Again, why?) (3) Let d be the common difference. Then a(n) = 1 + (d-1)*n a(10) / a(2) = a(34) / a(10) (Again, why?)