If x =(2/3)-⁴×(3/2)², find the value of.
a) x-² b) (x)-¹
Answers
Answer is 52
ANSWER
The coefficient of x¹⁰ in the expansion of (1+x)²(1+x²)³(1+x³)⁴ is equal to
the coefficient of x¹⁰ in the expansion of (1+x)²(1+x²)³(1+4x³+6x⁶+4x⁹)
We can ignore the last term in the expansion (1+x³)⁴, since its exponent is
greater than 10.
= Coefficient of x¹⁰ in the expansion of (1+x)²(1+x²)³
+4∗Coefficient of x⁷ in the expansion of (1+x)²(1+x²)³
+6∗Coefficient of x⁴ in the expansion of (1+x)²(1+x²)³
+4∗Coefficient of x in the expansion of (1+x)²(1+x²)³,
Coefficient of x¹⁰ in the expansion of (1+x)²(1+x²)³=0, since the highest degree term in the expansion is 8.
Coefficient of x⁷ in the expansion of (1+x)²(1+x²)³=
Coefficient of x⁷ in the expansion of (1+2∗x+x²)(1+x²)³
=2∗Coefficient of x⁶ in the expansion of (1+x²)³
=2∗1=2,
Coefficient of x⁴ in the expansion of (1+x)²(1+x²)³=
Coefficient of x⁴ in the expansion of (1+2∗x+x²)(1+x²)³
=1*Coefficient of x⁴ in the expansion of $$(1+x²)³ +
1*Coefficient of x in the expansion of (1+x²)³
=3+3=6
Coefficient of x in the expansion of (1+2∗x+x²)(1+x²)³
=2∗ constant in the expansion of (1+x²)³
=2,
Thus ,the coefficient of x¹⁰ in the expansion of (1+x)²(1+x²)³(1+x³)⁴
=0+4∗2+6∗6+4∗2
=52.