Expand (1 + )
2/3up to four terms for │ x │ < 1.
Answers
Answer:
Solution :-
Sum of coefficients of the first three terms of (x−
x
2
3
)
m
Now we know (- in the expansion (a+b)
n
,
the general term is T
r+1
=
n
C
r
a
n−r
b
r
In expansion (x−
x
2
3
)
m
, general term is T
r+1
=
m
C
r
x
m−r
(
x
2
−3
)
r
Now, first 3 terms are T
1
,T
2
and T
3
T
1
=
m
C
0
x
m
(
x
2
−3
)
0
=x
m
T
2
=
m
C
1
x
m−1
(
x
2
−3
)
1
=−3
m
C
1
x
m−3
=−3
3
C
1
T
3
=
m
C
2
(x)
m−2
(
x
2
−3
)
2
=
m
C
2
x
m−2
(
x
4
(−3)
2
)=9
m
C
2
.x
m−6
=9
m
C
2
Now given that, sum of first three = 559
1−3
m
C
1
+9
m
C
2
=559
⇒1−3×
1!(m−1)!
m!
+
(m−2)!2!
9×m!
=559
⇒1−3m+
2
9
m(m−1)=559
⇒2−6m+9m
2
−9m=559×2
⇒9m
2
−15m−1116=0
⇒(3m+31)(m−12)=0
⇒m=−
3
31
and m=12
∵m is natural number ∴m=12
Now, We need the coefficient of x
3
T
r+1
=
m
C
r
x
m−r
(
x
2
−3
)
r
=
12
C
r
x
12−r
(
x
2
−3
)
r
=
12
C
r
.(−3)
r
.x
12−3r
We need x
3
, ∴x
3
=x
12−3r
⇒r=3
Hence, the term containing x
3
T
r+1
=
12
C
r
.(−3)
r
.x
12−3r
T
3+1
=
12
C
3
.(−3)
3
x
12−9
=
6!3!
12!
×−27×x
3
T
4
=−5940x
3