InterviewSolution
Saved Bookmarks
InterviewSolution
| 1. |
Using the principle of mathematical induction, prove that ` 1/(1*2)+1/(2*3)+1/(3*4)+…+1/(n(n+1)) = n/((n+1)) `. |
|
Answer» Let the given statement be P(n). Then, ` P(n): 1/(1*2)+1/(2*3)+1/(3*4)+..+1/(n(n+1))= n/((n+1))`. Putting n = 1 in the given statement, we get LHS ` = 1/(1*2) = 1/2 and RHS = 1/((1+1)) = 1/2`. `:. ` LHS= RHS. Thus, P(1) is true. Let P(k) be true. Then, `P(k) : 1/(1*2)+1/(2*3)+1/(3*4) +...+ 1/(k(k+1)) = k/((k+1)) `. ...(i) Now, ` 1/(1*2)+1/(2*3)+1/(3*4)+...+1/(k(k+1)) + 1/((k+1)(k+2)) ` ` = {1/(1*2)+1/(2*3)+1/(3*4)+...+1/(k(k+1))} + 1/((k+1)(k+2)) ` ` = k/((k+1)) + 1/((k+1)(k+2))" "` [using (i)] ` =(k(k+2)+1)/((k+1)(k+2)) = ((k+1)^(2))/((k+1)(k+2)) = ((k+1))/((k+2))`. `:. " " P(k+1) : 1/(1*2)+1/(2*3)+1/(3*4)= 1/((k+1)(k+2)) = ((k+1))/((k+2)) .` Thus, P(k+1) is true, whenever P(k) is true. `:." "` P(1) is true and P(k+1) is true, whenever P(k) is true. Hence, by the principle of mathematical induction, we have ` 1/(1*2)+1/(2*3)+1/(3*4) +...+ 1/(n(n+1)) = n/(n+1)" for all " n in N`. |
|