find in ascending powers of x the first three terms of:
(2+x)(1+x)^8
Answers
Answer: S=(x
1/2
+
2x
1/4
1
)
n
=∑
n
C
r
.(x
1/2
)
n−r
.(
2x
1/4
1
)
r
=∑
n
C
r
.(
2
1
)
r
.x
(n−r)/2
.x
−r/4
⇒S=∑
r=0
n
n
C
r
(
2
1
)
r
x
(2n−3r)/4
We have to expand in decreasing powers of x.
∴S=
n
C
0
(
2
1
)
0
x
n/2
+
n
C
1
(
2
1
)
1
x
(2n−3)/4
+
n
C
2
(
2
1
)
2
x
(2n−6)/4
+......
First three terms form an A.P.
=
n
C
0
,
n
C
1
(
2
1
)&
n
C
2
(
2
1
)
2
from an A.P.
⇒
n
C
0
+
n
C
2
(
2
1
)
2
=2(
n
C
1
).
2
1
⇒1+
2
n(n−1)
×
4
1
=n
⇒(n−1)(n−8)=0
n
=1 (∵ there are at least 3 term)
⇒n=8
∴S=∑
r=0
8
8
C
r
(
2
1
)
2
x
4−3r/4
⇒4−
4
3r
=K; Where K∈I
Now of such values of r ranging from 0 to 8 are :→0,4 and 8
∴ There will be three such terms.
Hence, the answer is 3.
Answered By
toppr
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