Binomial Theorem: JEE Mains 2022 June 29 Shift 1

Medium· General and middle terms· +4 / −1
If the constant term in the expansion of (3x3−2x2+5x5)10\left(3x^3-2x^2+\dfrac{5}{x^5}\right)^{10} is 2k⋅l2^k\cdot l, where ll is an odd integer, then the value of kk is equal to :
Practise more Binomial Theorem PYQs