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Write the general term in the expansion ...

Write the general term in the expansion of `(x^2-y x)^(12),x!=0`

Text Solution

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We know that general term of expansion `(a+b)^n` is

`T_(r+1)=^nC_r(a)^(n-r). b^r`

For `(x^2-yx)`,

Putting `n=12`, `a=x^2` and `b=-yx`

`T_(r+1)=^12C_r(x^2)^(12-r). (-yx)^r`

`T_(r+1)=^12C_r(x)^(2(12-r)). (-1)^r.y^r.x^r`

`T_(r+1)=^12C_r(x)^(24-2r). (-1)^r.y^r.x^r`

`T_(r+1)=^12C_rx^(24-2r).x^r.y^r. (-1)^r`

`T_(r+1)=(-1)^r. " ^12C_r.x^(24-2r+r).y^r. `

`T_(r+1)=(-1)^r. " ^12C_r.x^(24-r).y^r `

Hence, General term = `(-1)^r. " ^12C_r.x^(24-r).y^r`
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