Write the general term in the expansion of (x2 – y x)12, x ≠ 0.

Asked by Abhisek | 1 year ago |  103

##### Solution :-

The general term Tr+1 in the binomial expansion is given by Tr+1 = r an-r br

Here n = 12, a= x2 and b = -y x

Substituting the values we get

Tn+1 =12Cr × x2(12-r) (-1)r yr xr

$$\dfrac{12!}{r!(12-r)!}\times x^{24-2r}-1^ry^rx^r$$

$$-1^r \dfrac{12!}{r!(12-r)!}x^{24-r}y^r$$

= -1r 12c.x24 –2r. yr

Answered by Pragya Singh | 1 year ago

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