If 'a' be the sum of the odd terms and 'b'
the sum of the even terms of the expansion
of (1 + x)
n, then (1 – x2)
n =
(A) a2 – b2 (B) a2 + b2
(C) b2 – a2 (D) None of these
Answers
Answer:
Step-by-step explanation:
Sum of odd terms in the expansion of (1+x)
n
will be 2
n−1
...(i)
Now
(1−x
2
)
n
=
n
C
0
−
n
C
1
x
2
+
n
C
2
x
4
+...(−1)
n
n
C
n
x
2n
...(n)
(1+x
2
)
n
=
n
C
0
+
n
C
1
x
2
+
n
C
2
x
4
+...
n
C
n
x
2n
...(m)
Subtracting n from m, we get
(1+x
2
)
n
−(1−x
2
)
n
=2[
n
C
1
x
2
+
n
C
3
x
6
+...]
Now let x=1.
Hence we get
2
n
−0=2[
n
C
1
+
n
C
3
+...]
Or
n
C
1
+
n
C
3
x
2
+
n
C
5
+...=2
n−1
Sum of odd terms in the expansion of (1+x)
n
will be 2
n−1
...(i)
Now
(1−x
2
)
n
=
n
C
0
−
n
C
1
x
2
+
n
C
2
x
4
+...(−1)
n
n
C
n
x
2n
...(n)
(1+x
2
)
n
=
n
C
0
+
n
C
1
x
2
+
n
C
2
x
4
+...
n
C
n
x
2n
...(m)
Subtracting n from m, we get
(1+x
2
)
n
−(1−x
2
)
n
=2[
n
C
1
x
2
+
n
C
3
x
6
+...]
Now let x=1.
Hence we get
2
n
−0=2[
n
C
1
+
n
C
3
+...]
Or
n
C
1
+
n
C
3
x
2
+
n
C
5
+...=2
n−1