Question

Use Pascal's triangle to expand the binomial.

Answer 1

Answer:

(d - 5)^6 = d^6 - 30*d^5 + 375*d^4 - 2500*d^3 + 9375*d^2 - 18750*d + 15625

Step-by-step explanation:

Suppose you have (x + y)^6, the numbers  in the 6th row of Pascal's triangle are the coefficients of the terms in a sixth order polynomial.

In the figure attached, Pascal's triangle is shown.

The coefficients are 1, 6, 15, 20, 15, 6,1. Then:

(x + y)^6 =  x^6 + 6*x^5*y + 15*x^4*y^2 + 20*x^3*y^3 + 15*x^2*y^4 + 6*x*y^5+ y^6

Replacing x = d and y = -5:

(d - 5)^6 =  d^6 + 6*d^5*(-5) + 15*d^4*(-5)^2 + 20*d^3*(-5)^3 + 15*d^2*(-5)^4 + 6*d*(-5)^5+ (-5)^6

(d - 5)^6 = d^6 - 30*d^5 + 375*d^4 - 2500*d^3 + 9375*d^2 - 18750*d + 15625

Answer 2

(d - 5y)⁶ = d⁶ - 30d⁵y + 375d⁴y² - 2500d³y³ + 9375d²y⁴ - 18750dy⁵ + 15625y⁶

Further explanation

The Problem:

Use Pascal's triangle to expand the binomial  (d - 5y)⁶.

The Process:

Look carefully at Pascal's triangle scheme in the attached picture.

Let us do a binomial expansion to:

$\boxed{ \ (a - b)^6 = a^6 - 6a^5b + 15a^4b^2 - 20a^3b^3 + 15a^2b^4 - 6ab^5 + b^6 \ }$, which comes from the following processing:

$\boxed{ \ (a - b)^6 = a^6 + 6a^5(-b) + 15a^4(-b)^2 + 20a^3(-b)^3 + 15a^2(-b)^4 - 6a(-b)^5 + (-b)^6 \ }$

Alright, see carefully how the expansion of this binomial expression.

$\boxed{ \ (d - 5y)^6 = ? \ }$

$\boxed{ \ (d - 5y)^6 = d^6 - 6d^5(5y) + 15d^4(5y)^2 - 20d^3(5y)^3 + 15d^2(5y)^4 - 6d(5y)^5 + (5y)^6 \ }$

$= d^6 - (5 \times 6)d^5y + (25 \times 15)d^4y^2 - (125 \times 20)d^3y^3 + (625 \times 15)d^2y^4 - (3125 \times 6)dy^5 + 15625y^6$

Thus, the expansion form of the binomial expression (d - 5y) ⁶ is $\boxed{\boxed{ \ d^6 - 30d^5y + 375d^4y^2 - 2500d^3y^3 + 9375d^2y^4 - 18750dy^5 + 15625y^6 \ }}$

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Alternative Steps

We can also use Newton's Binomial Expansion.

$\boxed{ \ (a + b)^n = \sum_{r = 0}^{n} \left(\begin{array}{ccc}n\\r\end{array}\right) a^{n- r}b^r \ }$

Remember this $\boxed{ \ \left(\begin{array}{ccc}n\\r\end{array}\right) = \frac{n!}{r!(n-r)!} \ }$

$\boxed{ \ (d - 5y)^6 = ? \ }$

$\boxed{ \ \sum_{r = 0}^{6} \left(\begin{array}{ccc}6\\0\end{array}\right) d^{6-0}(-5y)^6 \ }$

$\boxed{ \ = \left(\begin{array}{ccc}6\\0\end{array}\right) d^6(-5y)^0 \ }$ + $\boxed{ \ \left(\begin{array}{ccc}6\\1\end{array}\right) d^5(-5y)^1 \ }$ + $\boxed{ \ \left(\begin{array}{ccc}6\\2\end{array}\right) d^4(-5y)^2 \ }$ +  $\boxed{ \ \left(\begin{array}{ccc}6\\3\end{array}\right) d^3(-5y)^3 \ }$ +  $\boxed{ \ \left(\begin{array}{ccc}6\\4\end{array}\right) d^2(-5y)^4 \ }$ + $\boxed{ \ \left(\begin{array}{ccc}6\\5\end{array}\right) d^1(-5y)^5 \ }$ + $\boxed{ \ \left(\begin{array}{ccc}6\\6\end{array}\right) d^0(-5y)^6 \ }$

$\boxed{ \ = d^6 - 6d^5(5y) + 15d^4(25y^2) - 20d^3(125y^3) + 15d^2(625y^4) - 6d(3125y^5) + 15625y^6 \ }$

$\boxed{\boxed{ \ (d - 5y)^6 = d^6 - 30d^5y + 375d^4y^2 - 2500d^3y^3 + 9375d^2y^4 - 18750dy^5 + 15625y^6 \ }}$

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Notes

Pascal's triangle in common is a triangular array of binomial coefficients. It is undoubtedly entitled after the French mathematician Blaise Pascal, although other mathematicians studied it centuries before him in Italy, Germany, Persia, India, and China.

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Another example

$\boxed{ \ (2x - \frac{1}{2x})^3 = ? \ }$

$\boxed{ \ = 1 \cdot (2x)^3 + 3 \cdot (2x)^2 \Big(-\frac{1}{2x} \Big) + 3 \cdot 2x \Big(-\frac{1}{2x} \Big)^2 + 1 \cdot \Big(-\frac{1}{2x} \Big)^3 \ }$

$\boxed{ \ = (2x)^3 - 3 \cdot (2x)^2 \Big(\frac{1}{2x} \Big) + 3 \cdot 2x \Big(\frac{1}{2x} \Big)^2 - \Big(\frac{1}{2x} \Big)^3 \ }$

$\boxed{ \ = 8x^3 - 12x^2 \Big(\frac{1}{2x} \Big) + 6x \Big(\frac{1}{4x^2} \Big) - \Big(\frac{1}{8x^3} \Big) \ }$

$\boxed{ \ 8x^3 - 6x + \frac{3}{2x} - \frac{1}{8x^3} \ } \rightarrow \boxed{ \ = 8x^3 - 6x + \frac{3}{2}x^{-1} - \frac{1}{8}x^{-3} \ }$

• 2nd term in expansion of $\boxed{ \ (2x - \frac{1}{2x})^3 \ }$ is -6x.
• The coefficient of x⁻¹ in expansion of $\boxed{ \ (2x - \frac{1}{2x})^3 \ }$ is ³/₂.

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