If x∩=(1+1/n)^n , n is a natural number. . . then prove that (1) x∩ is an increasing sequence . . . (2) Prove that 2<x∩<3.
1(a) Find the term independent of x, (1+x)^p*(1+1/x)^q where p,q are integers. .
(b)If 1st three terms of (1+x)^n are in A.P. then prove that n(n-1)x2-4nx+1=0
(c)Prove by binomial theorem 13 divides (14^n-13^n-1)
11: x∩ = e simply & i don't think its an increasing sequence rather its decreasing and hence we have found the limit as e.
2: it can be proved easily as follows
e=1+1/1!+some positive quantity
hence e>2.
e=1+1/1!+1/2!+1/3!+.......<1+1/1+1/2+1/4+1/8+.......=1+1/(1-1/2)=3
hence e <3.
otherwise e=2.7182818 and its less than 3 and greater than 2.
12. a) (1+px+p(p-1)/2x^2+........) (1+q/x+q(q-1)/2*1/x^2+............)
independent term = 1+pq+pC2*qC2+............+qCp or pCq as q>p
or p>q.
b)first three terms are 1, nx, n (n-1)/2*x^2
and hence 2* nx= 1+ n (n-1)/2 *x^2.
c)14^n- 13^n -1= (1+13)^n- 13^n -1
by expanding binomial and subtracting 1& 13^n a number remains which is multiple of 13
1(c) write 14^n as(13+1)^n
Now expanding binomially=13^n+13^(n-1)+.....13n+1
now after subtracting every term contains 13 so it is divisible by 13...
1(a) it can simply be given by p+qCq..
simplify as (1+x)p+q/xq. now coefficient of xq in numerator is p+qCq.
66@ut10
x_n=\left(1+\dfrac{1}{n}\right)^n is not simply e. Rather, the number e is the limit of the sequence {xn}.
1yes sir i am wrong rather question should be asking limit of the given series then the answer would be e .
111) The inequality e^x\ge 1+x is well known.
Replace x by (x-1) to arrive at e^{x-1}\ge x .......(1).
Now let f(x)=\left(1+\frac{1}{x} \right)^x.
From which we have \frac{f'(x)}{f(x)}=ln\left(\frac{x+1}{x} \right)-1+\frac{x}{x+1}.
Obviously f(x)>0.
So let's bother with ln\left(\frac{x+1}{x} \right)-1+\frac{x}{x+1}.
Set \frac{x}{x+1}=k.
Needless to mention 0<k<1.
So we now need to show (k-1)-lnk>0 or more precisely, e^{k-1}>k....which is true by (1).
Thus the sequence is incresing.
11The form of 1st inequality actually used is e^{-x}\ge 1-x for positive x.
Which can be verified by bringing all terms to one side, then by showing it's increasing.
66To show that xn is increasing, probably it would be easier to consider the binomial expansion of (1+1/n)n.
\left(1+\dfrac{1}{n}\right)^n=1+1+\dfrac{1}{2!}\left(1-\dfrac{1}{n}\right)+\dfrac{1}{3!}\left(1-\dfrac{1}{n}\right)\left(1-\dfrac{2}{n}\right)+\ldots +\dfrac{1}{n!}\left(1-\dfrac{1}{n}\right)\cdots\left(1-\dfrac{n-1}{n}\right)
If now instead of n, we had n+1, first of all there will be one additional positive term, and secondly, each term (starting from the third one -- after the two 1's) will be greater than the corresponding term in the expansion of (1+1/n)n. Accordingly, xn is increasing.
Its quite obvious that xn is greater than 2. To prove that xn <3, we note that
x_n < 1+1+\dfrac{1}{2!}+\dfrac{1}{3!}+\ldots +\dfrac{1}{n!}
Further, we have
\dfrac{1}{3!}<\dfrac{1}{2^2},
\dfrac{1}{4!}<\dfrac{1}{2^3},
and so on, and finally
\dfrac{1}{n!}<\dfrac{1}{2^{n-1}}
Hence,
x_n<1+1+\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+\ldots +\dfrac{1}{2^{n-1}}<3-\dfrac{1}{2^{n-1}}<3
as required.
341not only easier but the right way.
When you use derivative of ex or ln x you are implicitly using the fact that the limit of the sequence is e!
1all the solutions posted by soumik and kaymant are correct. . . thanks. .
66@soumik
for #13: Its not okay. You a priori assumed the existence of limit. But the fact that this sequence has a limit follows from the fact that the given sequence is increasing and is bounded above.
11@ kaymant sir,
I was saying that to prove a function is increasing, we just find da derivative and prove f'(x)>0 na?
We don't bother abt limits der.
So why can't we do da same here if f(n)=an ?