1 ) Which bends in the direction of gravity but away from the light?
2) which bends toward light but away from the force of gravity
Answers
Answer:
Let (r + 1)th Term be the Term independent of x.
As We Know,
\begin{gathered} \large\bf T_{r+1}={}^{n} C_{r}. {( a)}^{n- r}.( {b)}^{r} \\ \end{gathered}
T
r+1
=
n
C
r
.(a)
n−r
.(b)
r
\begin{gathered} \large = \sf{}^{9} C_{r}. ( {{x}^{2})}^{9 - r} . \bigg(- \frac{ 1}{x} \bigg)^{r} \\ \end{gathered}
=
9
C
r
.(x
2
)
9−r
.(−
x
1
)
r
\begin{gathered}= \sf{}^{9} C_{r}.( - 1)^{r}.({{x)}^{2(9 - r)}} . \bigg(\frac{ 1}{x} \bigg)^{r} \\\end{gathered}
=
9
C
r
.(−1)
r
.(x)
2(9−r)
.(
x
1
)
r
\begin{gathered}= \sf{}^{9} C_{r}.( - 1)^{r}.({{x)}^{(18 - 2r)}} . \bigg(\frac{1}{x ^{r} } \bigg)\\\end{gathered}
=
9
C
r
.(−1)
r
.(x)
(18−2r)
.(
x
r
1
)
\begin{gathered}= \sf{}^{9} C_{r}.( - 1)^{r}.({{x)}^{(18 - 2r - r)}} \\\end{gathered}
=
9
C
r
.(−1)
r
.(x)
(18−2r−r)
\begin{gathered} \red{:\longmapsto\bf T_{r+1}={}^{9} C_{r}.( - 1)^{r}.({{x)}^{(18 - 3r)}}} \\\end{gathered}
:⟼T
r+1
=
9
C
r
.(−1)
r
.(x)
(18−3r)
★ As it is independent of x so power of x should be zero.
\begin{gathered}:\longmapsto18 - 3\text r = 0 \\ \end{gathered}
:⟼18−3r=0
\begin{gathered}:\longmapsto3\text r = 18 \\ \end{gathered}
:⟼3r=18
\begin{gathered}:\longmapsto\text r = \cancel\dfrac{18}{3} \\ \end{gathered}
:⟼r=
3
18
\purple{ \Large :\longmapsto \underline {\boxed{{\bf r = 6} }}}:⟼
r=6
★ According to our Assumption
(r + 1)th term is independent of x,