A few ques from algebra

You almost did it Karna.

Using : xT_n+1 = nT_n - (n+1)T_n+1

xT₂ = T₁ - 2T₂

xT₃ = 2T₂ - 3T₃

..... and so on

adding n such equations : x(T₂ + T₃ + .....+ T_n+1 ) = T₁ - (n+1)T_n+1

x(T₁ + T₂ + T₃ + .....+ T_n+1 ) = (1+x)T₁ - (n+1)T_n+1

xS_n+1 = 1 - (n+1)!(x+1)(x+2)....(x+n+1) (since T₁ = 1x+1 )

xS_n+1 = 1 - (1x+1) (2x+2) .... (n+1x+n+1)

all these bracketed fractions are smaller than 1 and if we multiply infinite such terms their limiting value is 0

xS_∞ = 1

S_∞ = 1/x

this sum can be done by method of difference

T_{n}=\frac{(n-1)!}{(x+1)(x+2)..........(x+n)}

T_{n+1}=\frac{n(n-1)!}{(x+1)(x+2)..........(x+n)(x+n+1)}

T_{n+1}=\frac{n T_{n}}{(x+n+1)}
hence

xT_{n+1}= n T_{n}-(n+1)T_{n+1}
hence answer is
\frac{1}{x}\lim_{n\rightarrow infinity}(\frac{n!}{(x+1)(x+2)(x+3).........(x+n)})

thanks for those answers

anyone doing Q4 ?

\sum{\frac{1}{\sqrt{k}+\sqrt{k+1}}} < \sum{\frac{1}{2\sqrt{k}}} <\sum{\frac{1}{\sqrt{k-1}+\sqrt{k}}}
{\frac{1(\sqrt{k})-\sqrt{k+1}}{\sqrt{k}+\sqrt{k+1}(\sqrt{k})-\sqrt{k+1})}}

-{\frac{(\sqrt{k})-\sqrt{k+1}}{1}}
\sum{{\frac{(\sqrt{k+1})-\sqrt{k}}{1}}}=9-1=8
hence answer is 16

Question. previously u had told the way to find generative function.
But I am unable to find the generative function of ax-1 + ax+1 = √2ax.

urs answer was Acos(Ï€x/4).
Please explain the way out sir.

Well known. The hint is:

\frac{1}{\sqrt k + \sqrt{k+1}} < \frac{1}{2\sqrt k} < \frac{1}{\sqrt{k-1} + \sqrt k}

Now can you nest the sum between two integers?

for ans (1)
http://targetiit.com/iit-jee-forum/posts/binomial-11560.html

ohk thanx for that link :)

@ hari sir can u plz explain a bit more on how u found out the function f(x) [1]

i dun hav any idea abt generating functions

i dont think such type of q can come in the exam......btw it was jus hit and trial with which those substis came.....its difficult for someone to strike des substis in the exam.....unlees u hav done it befor.....