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JEE Main20245 Apr 2024Morning ShiftMathematicsBinomial TheoremActual

If the constant term in the expansion of (1+2 x-3 x^3 ) ( 3 2 x^2- 1 3 x )^9 is p , then 108 p is equal to

Correct answer

0

Step-by-step solution

aligned & (1+2 x-3 x^3 ) ( 3 2 x^2- 1 3 x )^9 & General term m ( 3 2 x^2- 1 3 x )^9 & = ^9 C_r 3^ 9-2 r 2^ 9-r (-1)^r x^ 18-3 r aligned Put r=6 to get coeff. of x^0= ^9 C₆ 1 6^3 x^0= 7 18 x^0 Put r =7 to get coeff. of x ⁻³= ^9 C _ r 3⁻⁵ 2^2 (-1)^7 x ⁻³ aligned & =- ^9 C₇ 1 3^5 2^2 x⁻³= -1 27 x⁻³ & (1+2 x-3 x^3 ) ( 7 18 x^0- 1 27 x⁻³ ) & 7 18 + 3 27 = 7 18 + 1 9 = 7+2 18 = 9 18 = 1 2 & 108 1 2 =54 aligned

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