F.4 M2 Exam questions

(a) (1 + x)^k = 1 + kx + k(k - 1)x^2/2! + k(k - 1)(k - 2)x^3/3! + .......

2012-07-02 10:38:47 補充：
(b) The question looks like the expansion of e^y, so put (1 + x)^k = e^y, take log on both sides we get k ln (1 + x) = y.
So (1 + x)^k = e^y = 1 + y + y^2/2! + y^3/3! + ....... + y^r/r! = 1 + k ln (1 + x) + [k ln (1 +x)]^2/2! + ..... + [k ln (1 + x)]^r/r! + .....

2012-07-02 10:44:31 補充：
(c) We know integrating 1/(1 + x) = ln (1 + x). From part (a) put k = - 1, we get 1/(1 + x) = 1 - x + x^2 - x^3 + x^4 - x^5 + ........ Integrating both sides, ln (1 + x) = x - x^2/2 + x^3/3 - x^4/4 + ..... + C. When x = 0, LHS = ln (1 + 0) = ln 1 = 0. RHS = C, so C = 0. So answer is as integrated.

2012-07-02 20:39:15 補充：
To : Jj Have you learned integration? Please reply before I explain to you.

2012-07-03 05:49:55 補充：
Since you haven't learned integration, so suggest you skip this part until F.5.