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Prove by mathematical induction that 2^(2n-1) + 3^(2n-1) is divisible by 5 for all natural numbers n.

First check that this works for n=1:2^(2x1 - 1) + 3^(2x1 - 1) = 2^1 +3^1 = 5 (so true for n=1)Now we assume this to work for any n = k.Assumption: 2^(2k-1) + 3^(2k-1) = 5a, where a is some integer constant.N...
KI
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How can the integrating factor method be derived to give a solution to a differential equation?

Consider the general equationdy/dx + Py = Qwhere P and Q are functions of x.R (which will be introduced later) is also a function of x. So, all of a sudden, we are going to just state the product rule.We can...
JH
3128 Views

Prove by induction the sum of n consecutive positive integers is of the form n(n+1)/2.

As proof by induction goes we always have to show that it works for the base case, which in this case is the very first positive integer:1. So we show that the sum of the first number(s) is equal to one whic...
PS
9694 Views

solve 3sinh^2(2x) + 11sinh(2x) = 4 for x, giving your answer(s) in terms of the natural log.

3sinh^2(2x) + 11sinh(2x) - 4 = 0 --> (3sinh(2x) - 1)(sinh(2x) + 4) = 0 --> sinh(2x) = 1/3, sinh(2x) = -4(e^(2x) - e^(-2x))/2 = 1/3 --> e^(4x) -(2/3)e^(2x) - 1 = 0 --> e^(2x) = 1/3 + 2sqrt(5)/3 th...
WM
2931 Views

Write (1+2i) /(2-i) in form x+iy

(1+2i) (2+i)/(2+i)(2-i) = (2-2+4i+i)/5 = 5i/5 = i
RM
3353 Views