coefficients

2) Think of the problem geometrically. Let y₁ = √2-x₁² and y₂ = 9/x₂. Then (x₁, y₁) lies on the circle x² + y² = 2 while the point (x₂, y₂) lies on the rectangular hyperbola xy = 9. The given expression becomes (x₁ - x₂)² + (y₁ - y₂)² which is the square of the distance between the two points (x₁, y₁) and (x₂, y₂). So in effect we need to find the square of the minimum distance between the above two curves in the 1st quadrant. This will occur along the common normal. By the symmetry of the two curves the line y =x is the common normal. And so we need to find the distance square between the points where the line y = x meets the circle and the hyperbola.

For the circle, this happens at the point (1,1) while for the hyperbola its happens at the point (3,3). As such shortest distance between the two curves is 3√2 - √2 = 2√2. Hence the required minimum value is 8.

thankyou sir...

your first question is quite simple simply substitute n=100 n x=1.you get 0

\text{consider the expression } \\ (1-x)^n(1+\frac{1}{x}+\frac{1}{x^2}+\frac{1}{x^3}+\cdots+\frac{1}{x^{100}}) \\ \text{now look for the coefficient of }x^0 \text{ in it}

\text{coefficient of }x^0 \text{ in }(1-x)^n(\frac{1-\frac{1}{x^{101}}}{1-\frac{1}{x}}) \\ \text{coefficient of }x^{100} \text{ in }\left(1-x \right)^{n-1}\left(1-x^{101} \right)=\binom{n-1}{100}

dont know what is wrong with my logic ?

how did you cross check it shubodip, did you sum it up or what ??? :P

btw here is an alternative method

put x=1 in the expansion of (1-x)ⁿ

we have required expression = C0-C1+C2... +C(n-101) if n is odd

= -C0+ C1...+ C(n-101) if n is even

using (n,r)+ (n,r-1) = (n+1,r) we get the result.....:)