Question

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Explain how you would use the Binomial Theorem to expand (x + 3y)^7. Describe the pattern of both the coefficients and the exponents of each term in the expansion.

Answer 1

Refer to the attachment

Explanation:

You can ask me further questions if necessary

Answer 2

`x^7+21x^6y+189x^5y^2+945x^4y^3+2835x^3y^4+5103x^2y^5+5103xy^6+2187y^7`

Explanation:

The Binomial Theorem is a formula that allows you to expand the expression (a + b)ⁿ for any positive integer n.

`\boxed{\begin{array}{c}\underline{\sf Binomial\;Theorem}\\\\$\displaystyle (a+b)^n=\sum^(n)_(k=0)\binom{n}{k} a^(n-k)b^(k)$\\\\\\\textsf{where}\;\displaystyle \binom{n}{k} = (n!)/(k!(n-k)!)\\\end{array}}`

In the case of (x + 3y)⁷:

• a = x
• b = 3y
• n = 7

This means that:

• The exponents of the x term in the expansion decrease from 7 to 0 as k increases from 0 to 7.
• The exponents of the 3y term in the expansion increase from 0 to 7 as k increases from 0 to 7.

The coefficients of each term in the expansion are the binomial coefficients. These can be calculated using the formula given above, or obtained from row 7 of Pascal's triangle, which is 1, 7, 21, 35, 35, 21, 7, 1.

So, the expansion of (x + 3y)⁷ is:

`\begin{aligned}(x+3y)^7&=1 \cdot x^7 \cdot (3y)^0+\\&\;\;\;\;\;7\cdot x^6 \cdot (3y)^1+\\&\;\;\;\;\;21\cdot x^5 \cdot (3y)^2+\\&\;\;\;\;\;35\cdot x^4 \cdot (3y)^3+\\&\;\;\;\;\;35\cdot x^3 \cdot (3y)^4+\\&\;\;\;\;\;21\cdot x^2 \cdot (3y)^5+\\&\;\;\;\;\;7\cdot x^1 \cdot (3y)^6+\\&\;\;\;\;\;1\cdot x^0 \cdot (3y)^7\end{aligned}`

This simplifies to:

`\boxed{x^7+21x^6y+189x^5y^2+945x^4y^3+2835x^3y^4+5103x^2y^5+5103xy^6+2187y^7}`