66The first one is e-2
Q1. \lim_{n\rightarrow \infty }\sum_{k=0}^{n}{\frac{^nC_{k}}{n^k(k+3)}}
Q2. If the roots of the equation ax2+bx+c=0, a>0 are positive and are p and q (p>q) then
\lim_{x\rightarrow 1/p^+}\sqrt{\frac{1-cos[2(cx^2+bx+c)]}{(1-px)^2}}
edit: the second one is cx2+bx+a in the cos()
-
UP 0 DOWN 0 1 14
14 Answers
66have a class right now...will come back and post the solution..if no one has done it by then.
21first one i hav done before.... its like this
\lim_{n\rightarrow \infty }\sum_{k=0}^{n}\frac{{^{n}C_{k}}}{n^{k}(k+3)}= \lim_{n\rightarrow \infty }\sum_{k=0}^{n}\frac{{^{n}C_{k}}}{n^{k}}.\int_{0}^{1}{x^{k+2}dx}=\int_{0}^{1} \lim_{n\rightarrow \infty }\sum_{k=0}^{n}\frac{{^{n}C_{k}}}{n^{k}}.{x^{k+2}dx}= \int_{0}^{1}x^{2}\left[\lim_{n\rightarrow \infty}\sum_{k=0}^{n}{^{n}C_{k}(\frac{x}{n})^{k}}\right]dx = \int_{0}^{1}x^{2}\left\left(\lim_{n\rightarrow \infty }(1+ \frac{x}{n})^{n})dx=\int_{0}^{1}{x^{2}.e^{x}}dx
last step we wrote using the fact tat \lim_{n\rightarrow \infty }(1+ \frac{a}{x})^{x}=e^{a} (a typo its x→∞)
11Since p and q are the roots of ax2+bx+c = 0
ax2+bx+c = 0 is reciprocal of cx2+bx+a = 0
Therefore, cx2 + bx + a = c(c- 1/p) (x - 1/q) = 0
Therefore, cx2+ bx+a = 0 has roots 1/p , 1/q
lim x→1/p+ √[1 - cos(cx2 + bx+a)][2 ( 1- px) 2]
= lim x→1/p+ √[2 sin 2(cx2 + bx+a)/2][2 ( 1- px) 2]
= lim x→1/p+ √[2 sin 2(c(x - 1/p) ( x - 1/q)/2)][2 ( 1- px) 2]
= lim x→1/p+ √[2 sin 2(c(x - 1/p) ( x - 1/q)/2)][(c(x- 1/p) (x- 1/q)/2] 2 X c2 (x- 1/p)2 (x- 1/q)24 (2p2) (x-1/p)2
On simplifying , we get,
= 1/2 (1/p -1/q) |c/p|
= (c / 2p) [ q-p / pq]
21\frac{1}{k+3}=\int_{0}^{1}{x^{k+2}}dx
is der any prob understanding this......
21dosrey step mey kya nahi samaj nahi aaya bhai[11].........i think after fisrt step every step is very clear.....which a person of ur level can understand easily
23hey asish eragon has done it nicely
he has taken that lim inside since its a constant
and he wrote xk+2 = xk.x2 and took that xk with nk