# Using the binomial theorem to find term with x^n?

How do you find the term that contains x^j (using the formula (n choose n-j)(x^j)(a^n-j)) when a has an x in the denominator? (Ex. (x+1/x)^6, find the term with x^3)

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- MathmomLv 73 years agoFavorite Answer
(x + 1/x)^6 = Σ [k=0 to 6] C(6,k) x^(6−k) (1/x)^k

To find term with x^3, then find k so that exponent of x is 3 in

x^(6−k) (1/x)^6

x^(6−k) (1/x)^k = x^3

x^(6−k) / x^k = x^3

x^(6−k−k) = x^3

6−2k = 3

No integer value of k gives us an exponent of 3.

So (x + 1/x)^6 has no term with x^3

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Let's find term with x^2 instead

x^(6−k) (1/x)^k = x^2

x^(6−k) / x^k = x^2

x^(6−k−k) = x^2

6−2k = 2

k = 2

Term: C(6,2) x^(6−2) (1/x)^2 = 15 * x^4 / x^2 = 15x^2

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