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Sagot :
Vedic Math Techniques informal methods such as chunking, grids, number lines and visual method :)
Well I really don't get the question but I'll answer it according to my under standing
let say we got this binomial equation in terms of x and y
[tex](x+y)^n=0[/tex]
first the number of terms is the expansion is n+1
the first term is [tex]x^n[/tex] and the last term is [tex]y^n[/tex]
the exponent of [tex]x[/tex] descends linearly from n to 0
the exponent of [tex]y[/tex] descends linearly from 0 to n
the sum of the exponents of a and b in any of the terms is equal to n,
the coefficient of the second term and the second to the last term is n,
so
[tex](x+y)^n=(x^ny^0+n x^{n-1} y^1+ \frac{n}{2} x^{n-2} y^2+...+ \frac{n}{n-1}x^1 y^{n-1} + \frac{n}{n} x^{0} y^{n} )[/tex]
example:
[tex](x+y)^3[/tex]
in this example n = 3
the number of therms is 3 + 1 = 4
so there are 4 terms in this example
[tex] (x+y)^{3} = [/tex][tex]( x^{3}y^0 +3 x^{3-1} y^{1} +3x y^{3-1} + \frac{3}{3}y^3) [/tex]
[tex] (x+y)^{3} = x^{3} y^{0} +3 x^{2} y^{1} +3 x^{1} y^{2} + x^{0} y^{3} [/tex]
see the pattern in which the exponent of x is decreasing from 3 to 0
and the pattern in which the exponent of y is increasing from 0 to 3
this is the pattern that I knew in multiplying binomial algebraic expression
don't know if this is the answer you are looking for.
hit thank you if this helps you..
let say we got this binomial equation in terms of x and y
[tex](x+y)^n=0[/tex]
first the number of terms is the expansion is n+1
the first term is [tex]x^n[/tex] and the last term is [tex]y^n[/tex]
the exponent of [tex]x[/tex] descends linearly from n to 0
the exponent of [tex]y[/tex] descends linearly from 0 to n
the sum of the exponents of a and b in any of the terms is equal to n,
the coefficient of the second term and the second to the last term is n,
so
[tex](x+y)^n=(x^ny^0+n x^{n-1} y^1+ \frac{n}{2} x^{n-2} y^2+...+ \frac{n}{n-1}x^1 y^{n-1} + \frac{n}{n} x^{0} y^{n} )[/tex]
example:
[tex](x+y)^3[/tex]
in this example n = 3
the number of therms is 3 + 1 = 4
so there are 4 terms in this example
[tex] (x+y)^{3} = [/tex][tex]( x^{3}y^0 +3 x^{3-1} y^{1} +3x y^{3-1} + \frac{3}{3}y^3) [/tex]
[tex] (x+y)^{3} = x^{3} y^{0} +3 x^{2} y^{1} +3 x^{1} y^{2} + x^{0} y^{3} [/tex]
see the pattern in which the exponent of x is decreasing from 3 to 0
and the pattern in which the exponent of y is increasing from 0 to 3
this is the pattern that I knew in multiplying binomial algebraic expression
don't know if this is the answer you are looking for.
hit thank you if this helps you..
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