JEE Advanced Binomial Theorem 1

If (x+a)100=t0+t1++t100(x+a)^{100}=t_0+t_1+\cdots+t_{100} where tr=(nr)xnrart_r=\dbinom{n}{r}x^{n-r}a^r, then 2xdx(r=050(1)rt2r)2+(r=049(1)rt2r+1)2=C+αβ(xη+aη)γ,\displaystyle \int\dfrac{2xdx}{\left (\displaystyle\sum_{r=0}^{50}(-1)^rt_{2r}\right )^2+\left (\displaystyle\sum_{r=0}^{49}(-1)^rt_{2r+1}\right )^2}=C+\dfrac{\alpha}{\beta(x^\eta+a^\eta)^\gamma},

If βη>0,GCD(α,β)=1\beta\eta>0, GCD(|\alpha|,|\beta|)=1, Calculate α+β+η+γ\alpha+\beta+\eta+\gamma

This is a part of My Picks for JEE Advanced 2
×

Problem Loading...

Note Loading...

Set Loading...