Find the value of x - y 4 using Pascal-s triangle- A- x4-3 x3 y-5 x2 y2-3 x y3-y4B- x4-2 x3 y-6 x2 y2-2 x y3-y4C- x4-4 x3 y-6 x2 y2-4 x y3-y4D- x4-4 x3 y-4 x2 y2-4 x y3-y4

The correct option is C x^4 +4x^3y + 6x^2y^2+ 4xy^3 + y^4We know that the value of (x+y)^n for any value of n ( gt;0) can be obtained using the Pascal's tria ...

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Byju's Answer

Standard VIII

Mathematics

Multiplication of Any Polynomial

Find the valu...

Question

Find the value of $(x + y)^{4}$ using Pascal's triangle.

x^{4} + 3 x^{3} y + 5 x^{2} y^{2} + 3 x y^{3} + y^{4}

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x^{4} + 2 x^{3} y + 6 x^{2} y^{2} + 2 x y^{3} + y^{4}

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x^{4} + 4 x^{3} y + 6 x^{2} y^{2} + 4 x y^{3} + y^{4}

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x^{4} + 4 x^{3} y + 4 x^{2} y^{2} + 4 x y^{3} + y^{4}

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Solution

The correct option is C $x^{4} + 4 x^{3} y + 6 x^{2} y^{2} + 4 x y^{3} + y^{4}$
We know that the value of $(x + y)^{n}$ for any value of n (>0) can be obtained using the Pascal's triangle.

The coefficients for the expression $(x + y)^{n}$ are as follows:

$(x + y)^{4} = 1 x^{4} y^{0} + 4 x^{3} y^{1} + 6 x^{2} y^{2} + 4 x^{1} y^{3} + 1 x^{0} y^{4} = x^{4} + 4 x^{3} y + 6 x^{2} y^{2} + 4 x y^{3} + y^{4}$

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Find the value of (x+y)4 using Pascal's triangle.

The correct option is C x4+4x3y+6x2y2+4xy3+y4We know that the value of (x+y)n for any value of n (>0) can be obtained using the Pascal's triangle. The coefficients for the expression (x+y)n are as follows: (x+y)4=1x4y0+4x3y1+6x2y2+4x1y3+1x0y4 =x4+4x3y+6x2y2+4xy3+y4

Find the value of $(x + y)^{4}$ using Pascal's triangle.