The term independent of x in the
expansion of (ax+b/x)^14 is
a)14c5a^9b^5
b)14c6a^8b^6
c)14c7a^7b^7
d)14c8a^6b^8
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Answers
Given:
The expansion of (ax+b/x)^14
To Find:
The term independent of x in the expansion of (ax+b/x)^14.
Solution:
The jth term in the binomial expansion of the term =>
Therefore in the given question, let jth term be independent of x.
Each term in the expansion of (ax+b/x)^14 can be written in the form as jth term =>
=
For the terms to be independent of x , and
should cancel each other.
Therefore,
- 2j = 14
- j = 7
Hence the 7th term will be independent of x.
The term independent of x in the expansion of (ax+b/x)^14 is .
Step-by-step explanation:
The expansion of (ax+b/x)^14
To Find:
The term independent of x in the expansion of (ax+b/x)^14.
Solution:
The jth term in the binomial expansion of the term (a+b)^{n}(a+b)
n
=>
nC_j a^{j} b^{n-j}nC
j
a
j
b
n−j
Therefore in the given question, let jth term be independent of x.
Each term in the expansion of (ax+b/x)^14 can be written in the form as jth term =>
14C_j (ax)^{j}(\frac{b}{x} )^{14-j}14C
j
(ax)
j
(
x
b
)
14−j
= 14C_j a^{j} b^{14-j}\frac{ x^{2j } }{x^{14} }14C
j
a
j
b
14−j
x
14
x
2j
For the terms to be independent of x , x^{2j}x
2j
and x^{14}x
14
should cancel each other.
Therefore,
2j = 14
j = 7
Hence the 7th term will be independent of x.
The term independent of x in the expansion of (ax+b/x)^14 is 14C_7a^{7} b^{7}14C
7
a
7
b
7
.