Write the general term in the expansion of (x2 – y)6

Asked by Abhisek | 1 year ago |  90

##### Solution :-

The general term Tr+1 in the binomial expansion is given by

Tr+1 = r an-r br…….. (i)

Here a = x2 , n = 6 and b = -y

Putting values in (i)

Tr+1 = 6Cr x 2(6-r) (-1)r yr

$$\dfrac{6!}{r!(6-r)!}\times x^{12-2r}\times (-1)^r\times y^r$$

$$(-1)^r \dfrac{6!}{r!(6-r)!}\times x^{12-2r}\times y^r$$

= -1r 6c.x12 – 2r. yr

Answered by Pragya Singh | 1 year ago

### Related Questions

#### Find the term independent of x in the expansion of (3/2 x2 – 1/3x)9

Find the term independent of x in the expansion of $$(\dfrac{3}{2x^2} – \dfrac{1}{3x})^9$$

#### Find the middle term in the expansion of (x – 1/x)2n+1

Find the middle term in the expansion of $$(x-\dfrac{ 1}{x})^{2n+1}$$

#### Find the middle term in the expansion of (1 + 3x + 3x2 + x3)2n

Find the middle term in the expansion of (1 + 3x + 3x2 + x3)2n

Find the middle term in the expansion of $$(\dfrac{x}{a} – \dfrac{a}{x})^{10}$$