binomials and problems regarding binomials

(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)

1: 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.

2. a) (1+px+p(p-1)/2x^2+........) (1+q/x+q(q-1)/2*1/x^2+............)
independent term = 1+pq+pC₂*qC₂+............+qC_p or pC_q 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

(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...

(a) it can simply be given by ^p+qC_q..
simplify as (1+x)^p+q/x^q. now coefficient of x^q in numerator is ^p+qC_q.

@ut10
x_n=\left(1+\dfrac{1}{n}\right)^n is not simply e. Rather, the number e is the limit of the sequence {x_n}.

yes sir i am wrong rather question should be asking limit of the given series then the answer would be e .

1) 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.

The 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.

To show that x_n 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, x_n is increasing.

Its quite obvious that x_n is greater than 2. To prove that x_n <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.

not only easier but the right way.

When you use derivative of e^x or ln x you are implicitly using the fact that the limit of the sequence is e!

But tht's ok. Limit in this case is e - as said in #7.

all the solutions posted by soumik and kaymant are correct. . . thanks. .

@ 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)=a_n ?

That's okay..