Question #129750
Find the coefficient x to the power 4 * y cube & number of terms in the expansion of (3x – 10y)11
1
Expert's answer
2020-08-17T15:39:26-0400

Expansion of (3x10y)11(3x – 10y)^{11} =C(11,0)×(3x)11×(10y)0+C(11,1)×(3x)10×(10y)1+C(11,2)×(3x)9×(10y)2+...+C(11,10)×(3x)1×(10y)10+C(11,11)×(3x)0×(10y)11= C(11,0)\times (3x)^{11} \times (-10y)^0 + C(11,1) \times (3x)^{10} \times (-10y)^1+ C(11,2) \times (3x)^{9} \times (-10y)^2 + ... + C(11,10) \times (3x)^{1} \times (-10y)^{10} + C(11,11) \times (3x)^0 \times (-10y)^{11}

=311x11(11×310×101)x10y1+(55×39×102)x9y2+...+(11×31×1010)x1y101011y11= 3^{11} x^{11} -(11 \times 3^{10} \times 10^1) x^{10} y^1 + (55 \times 3^9 \times 10^2) x^9 y^2 + ... + (11 \times 3^{1} \times 10^{10}) x^{1} y^{10} - 10^{11} y^{11}

General term is Tr+1=C(11,r)(3x)11r(10y)rT_{r+1} = C(11,r) (3x)^{11-r} (-10y)^{r} .

Hence, number of terms are 12.

Term containing y3y^3 appears when r=3r = 3 i.e. fourth term.

So, T4=C(11,3)(3x)8(10y)3=(165000×38)x8y3T_4 = C(11,3) (3x)^8 (-10y)^3 = (165000 \times 3^8) x^8 y^3 .

This term is not of x4y3x^4 y^3 form, hence coefficient of x4y3x^4 y^3 is zero.


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