The term independent of x in the expansion of (1-x^2+3x^3)(52x^3-15x^2)^11, x0, is:

Question

The term independent of $x$ in the expansion of $(1-x^2+3x^3)\left(\dfrac{5}{2}x^3-\dfrac{1}{5x^2}ight)^{11}$, $x
e0$, is:

Accepted Answer

General term of $\left(\dfrac52x^3-\dfrac1{5x^2}\right)^{11}$: $T_{r+1}=\binom{11}{r}\left(\dfrac52\right)^{11-r}\left(-\dfrac15\right)^r x^{33-5r}$

Multiplying by $1,\ -x^2,\ 3x^3$ from the first factor, we need $33-5r\in\{0,-2,-3\}$. Only $33-5r=-2$ gives integer $r=7$:

$$-\binom{11}{7}\left(\frac52\right)^4\left(-\frac15\right)^7 = \binom{11}{7}\frac{5^4}{2^4}\cdot\frac{1}{5^7}=\frac{11\cdot10\cdot9\cdot8}{4\cdot3\cdot2\cdot1}\cdot\frac{1}{2^4\cdot5^3}=\frac{330}{2000}=\frac{33}{200}$$

Solutions · 1