The term independent of x in the expansion of (x+1/x2?)6 is
Question:

The term independent of xx in the expansion of (x+1x2)6\left(x+\frac{1}{x^{2}}\right)^{6} is

Updated On: May 18, 2024
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The Correct Option is B

Solution and Explanation

The general term of (x+1x2)6\left(x+\frac{1}{x^{2}}\right)^{6} is
Tr+1=6Crxr(1x2)6rT_{r+1}={ }^{6} C_{r} x^{r}\left(\frac{1}{x^{2}}\right)^{6-r}
=6Crxr12+2r={ }^{6} C_{r} x^{r-12+2 r}
For independent of xx,
r12+2r=0r-12+2 r=0
3r=12\Rightarrow 3 r=12
r=4\Rightarrow r=4
\therefore Required term =6C4=6×52×1=15={ }^{6} C_{4}=\frac{6 \times 5}{2 \times 1}=15
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Concepts Used:

Binomial Theorem

The binomial theorem formula is used in the expansion of any power of a binomial in the form of a series. The binomial theorem formula is 

Properties of Binomial Theorem

  • The number of coefficients in the binomial expansion of (x + y)n is equal to (n + 1).
  • There are (n+1) terms in the expansion of (x+y)n.
  • The first and the last terms are xn and yn respectively.
  • From the beginning of the expansion, the powers of x, decrease from n up to 0, and the powers of a, increase from 0 up to n.
  • The binomial coefficients in the expansion are arranged in an array, which is called Pascal's triangle. This pattern developed is summed up by the binomial theorem formula.