CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon


Question

The general term in the expansion of $$ \displaystyle \left ( 1+3x-2x^{3} \right ) $$ is


A
10!r1!r2!r3!(1)r33r22r3xr2+3r3
loader
B
10!r1!r2!r3!(1)r23r32r3xr2+3r3
loader
C
r1+r2+r3=10 10!r1!r2!r3!3r22r3(1)r3xr2+3r3
loader
D
None of these
loader

Solution

The correct option is C $$ \displaystyle \sum_{r_{1}+r_{2}+r_{3}=10}^{} $$ $$ \displaystyle \frac{10!}{r_{1}!r_{2}!r_{3}!}3^{r_{2}}2^{r_{3}}\left ( -1 \right )^{r_{3}}x^{r_{2}+3r_{3}} $$
$$(x_{1}+x_{2}+\cdots+x_{k})^{n}=\displaystyle\sum_{r_{1}+r_{2}+r_{3}+\cdots+r_{k}=n} \dfrac{n!}{r_{1}!r_{2}!r_{3}!..r_{k}!}x_{1}^{r_{1}}x_{2}^{r_{2}}...x_{n}^{r_{n}}$$
$$(1+3{x}-2{x^{3}})^{10}=\displaystyle\sum_{r_{1}+r_{2}+r_{3}=10}\dfrac{10!}{r_{1}!r_{2}!r_{3}!}(1)^{r_{1}}(3{x})^{r_{2}}(-2{x^{3}})^{r_{3}}$$
General term is $$\dfrac{10!}{r_{1}!r_{2}!r_{3}!}3^{r_{2}}2^{r_{3}}(-1)^{r_{3}}x^{r_{2}+3r_{3}}$$

Maths

Suggest Corrections
thumbs-up
 
0


similar_icon
Similar questions
View More


similar_icon
People also searched for
View More



footer-image