Show that nC1 + nC5 + nC9 + ... = 1/2 (2n-1 + 2n/2 sin (n pi/4))
With the calculus of residues show that (2n)! cos2n 6 do = 1720 (91) 2 (2n – 1)!! -, =- " n= 0, 1, 2, ... (2n)!! Hint. cos 0 = (eie +e-10)/2 = (2+z-1)/2, 1z| = 1.
Prove that P2n(0)= (-1)n ((2n-1)!!/(2n)!!) using the generation function and a binomial expansion. Show that (sqrt(pi)(4n-1)/(2gamma(n+1)gamma(3/2-n))=(-1)n-1((2n-3)!!/(2n-2)!!)(4n-1)/2n
9. Using the Binomial Theorem, show that Σk㈡-n 2n-1
Diagrammatically show the following steps in division of a 2n = 4 cell with 1 autosome, and 1 X-chromosome, and 1 Y-chromosome in: A.) Anaphase I of meiosis
1. Show that, for every n > 1: n ka n(n + 1)(2n +1) 6 k=1
Test the series below for convergence using the Ratio Test. 10" 2n! The limit of the ratio test simplifies to lim f(n) where f(n) = Preview Preview The limit is: (enter oo for infinity if needed) Based on this, the series Converges Message instructor about this question
Q1
Show from first principles that 2n2 + 2n +1 - 4n2 +3 1 lim n+
. Show that, for every η Σ1: Τη Σ2, η(η + 1)(2n + 1) 6 k=1
Show that the sequence is Cauchy using the definition of Cauchy se- quences. Sn 2n +1 n +4