Statement => Binomial theorem states that for any given positive integer n, the expression of the nth power of the sum of any two numbers a and b may take place as the sum of n+1 terms of the particular form.
State and prove Binomial theorem for any positive integer n.
Easy
Solution
verified
Verified by Pritam
The proof is obtained by principal of mathamatical induction
let the given statement be
P(n):(a+b)
n
=
n
C
0
a
n
+
n
C
1
a
n−1
b+
n
C
2
a
n−2
b
2
+.....+
n
C
n−1
a.b
n−1
+
n
C
n
b
n
for n=1
p(1):(a+b)
1
=
1
C
0
a
1
+
1
C
1
b
1
=a+b
thus p(1) is true.
suppose p(k) is true for some positive integer k,
(a+b)
k
=
k
C
0
a
k
+
k
C
1
a
k−1
b+
k
C
2
a
k−2
b
2
+.....+
k
C
k
b
k
we shall proove that P(k+1) is also true,
(a+b)
k+1
=
k+1
C
0
a
k+1
+
k+1
C
1
a
k
b+
k+1
C
2
a
k−1
b
2
+...+
k+1
C
k+1
b
k+1
now
(a+b)
k+1
=(a+b)(a+b)
k
=(a+b)(
k
C
0
a
k
+
k
C
1
a
k−1
b+
k
C
2
a
k−2
b
2
+...+
k
C
k−1
ab
k−1
+
k
C
k
b
k
)......(1)
=
k
C
0
a
k+1
+
k
C
1
a
k
b+
k
C
2
a
k−1
b
2
+...+
k
C
k−1
a
2
b
k−1
+
k
C
k
ab
k
+
k
C
0
a
k
b+
k
C
1
a
k−1
b
2
+
k
C
2
a
k−2
b
3
+...+
k
C
k−1
ab
k
+
k
C
k
b
k+1
=
k
C
0
a
k+1
+(
k
C
1
+
k
C
0
)a
k
b+(
k
C
2
+
k
C
1
)a
k−1
b
2
+...+(
k
C
k
+
k
C
k−1
)ab
k
+
k
C
k
b
k+1
=
k+1
C
0
a
k+1
+
k+1
C
1
a
k
b+
k+1
C
2
a
k−1
b
2
+...+
k+1
C
k
ab
k
+
k+1
C
k+1
b
k+1
(using
k+1
C
0
=1,
k
C
r
+
k
C
r−1
=
k+1
C
r
,
k
C
k
=1=
k+1
C
k+1
)
.
Thus, it has been proved that P (k + 1) is true whenever P(k) is true. Therefore, by principle of mathematical induction, P(n) is true for every positive integer n.
Answers & Comments
Verified answer
Statement => Binomial theorem states that for any given positive integer n, the expression of the nth power of the sum of any two numbers a and b may take place as the sum of n+1 terms of the particular form.
Answer:
State and prove Binomial theorem for any positive integer n.
Easy
Solution
verified
Verified by Pritam
The proof is obtained by principal of mathamatical induction
let the given statement be
P(n):(a+b)
n
=
n
C
0
a
n
+
n
C
1
a
n−1
b+
n
C
2
a
n−2
b
2
+.....+
n
C
n−1
a.b
n−1
+
n
C
n
b
n
for n=1
p(1):(a+b)
1
=
1
C
0
a
1
+
1
C
1
b
1
=a+b
thus p(1) is true.
suppose p(k) is true for some positive integer k,
(a+b)
k
=
k
C
0
a
k
+
k
C
1
a
k−1
b+
k
C
2
a
k−2
b
2
+.....+
k
C
k
b
k
we shall proove that P(k+1) is also true,
(a+b)
k+1
=
k+1
C
0
a
k+1
+
k+1
C
1
a
k
b+
k+1
C
2
a
k−1
b
2
+...+
k+1
C
k+1
b
k+1
now
(a+b)
k+1
=(a+b)(a+b)
k
=(a+b)(
k
C
0
a
k
+
k
C
1
a
k−1
b+
k
C
2
a
k−2
b
2
+...+
k
C
k−1
ab
k−1
+
k
C
k
b
k
)......(1)
=
k
C
0
a
k+1
+
k
C
1
a
k
b+
k
C
2
a
k−1
b
2
+...+
k
C
k−1
a
2
b
k−1
+
k
C
k
ab
k
+
k
C
0
a
k
b+
k
C
1
a
k−1
b
2
+
k
C
2
a
k−2
b
3
+...+
k
C
k−1
ab
k
+
k
C
k
b
k+1
=
k
C
0
a
k+1
+(
k
C
1
+
k
C
0
)a
k
b+(
k
C
2
+
k
C
1
)a
k−1
b
2
+...+(
k
C
k
+
k
C
k−1
)ab
k
+
k
C
k
b
k+1
=
k+1
C
0
a
k+1
+
k+1
C
1
a
k
b+
k+1
C
2
a
k−1
b
2
+...+
k+1
C
k
ab
k
+
k+1
C
k+1
b
k+1
(using
k+1
C
0
=1,
k
C
r
+
k
C
r−1
=
k+1
C
r
,
k
C
k
=1=
k+1
C
k+1
)
.
Thus, it has been proved that P (k + 1) is true whenever P(k) is true. Therefore, by principle of mathematical induction, P(n) is true for every positive integer n.