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Sequences & Series

Binomial Theorem & Pascal's Triangle

Introduction

Consider the 3rd power of \((a + b)\):

\[ (a+b)^3 = (a+b)(a+b)(a+b) \]

On multiplying out and simplifying like terms we come up with the results:

\[ (a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3 \]

Note that each term is a combination of \(a\) and \(b\) and the sum of the exponents is equal to 3 for each term. The degree of each term is 3.

Thus, if we expand \((a + b)^4 = \_a^4 + \_a^3b + \_a^2b^2 + \_ab^3 + \_b^4\) the question is, what are the coefficients? These coefficients are called binomial coefficients. In the expansion of \((a + b)^3\) the binomial coefficients are 1   3   3   1. Note that there is a "left to right" and "right to left" symmetry to the numbers. These coefficients can be obtained by the use of Pascal's Triangle.

Pascal's Triangle
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This triangular array is called Pascal's Triangle. Each row gives the combinatorial numbers, which are the binomial coefficients. That is, the row 1   2   1 is the combinatorial numbers \({}^{2}C_r\), which are the coefficients of \((a + b)^2\). The next row, 1   3   3   1, is the coefficients of \((a + b)^3\); and so on.

To construct the triangle, write 1, and below it write 1   1. Begin and end each successive row with 1. To construct the intervening numbers, add the two numbers immediately above.

To construct the next row, begin it with 1, and add the two numbers immediately above: 1 + 2. Write 3. Again, add the two numbers immediately above: 2 + 1 = 3. Finish the row with 1.

There are instances where the expansion of the binomial is so large that Pascal's Triangle is not advisable to be used. An easier way to expand a binomial raised to a certain power is through the binomial theorem. It is finding the solution to the problem of the binomial coefficients without actually multiplying out. The theorem is given as:

\[ (a+b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k \]

Which can be expanded as:

\[ (a+b)^n = \binom{n}{0}(a)^n + \binom{n}{1}(a)^{n-1}(b) + \binom{n}{2}(a)^{n-2}(b)^2 + \cdots + \binom{n}{n}a^0(b)^n \]

Remember that \(\binom{n}{k}\) is another way of writing combination \(= {}^{n}C_r = \frac{n!}{(n-r)!\,r!}\).

The Binomial Theorem also applies to a binomial with literal coefficients. It is given as:

\[ (a+bx)^n = \binom{n}{0}(a)^n + \binom{n}{1}(a)^{n-1}(bx) + \binom{n}{2}(a)^{n-2}(bx)^2 + \cdots + \binom{n}{n}(bx)^n \]

How to solve different types of problems

Proving an expression when terms above a certain threshold can be ignored

Question

If \(x\) is too small so that terms of \(x^3\) and higher can be ignored, show that \((4-x)(1-2x)^6 = 4 - 49x + 252x^2\)

Solution
  1. Step 1 — Write out the expansion

    We use Pascal's Triangle in the expansion of \((1-2x)^6\). The index of \((1-2x)^6\) is \(6\), so we look on the 7th line of Pascal's Triangle. The coefficients are \(1, 6, 15, 20, 15, 6, 1\):

    \[ (4-x)(1-2x)^6 = (4-x)\left[ 1(1)^6 + 6(1)^5(-2x) + 15(1)^4(-2x)^2 + 20(1)^3(-2x)^3 + 15(1)^2(-2x)^4 + 6(1)(-2x)^5 + 1(-2x)^6 \right] \]
  2. Step 2 — Simplify the product
    \[ = 4 - 48x + 240x^2 - 640x^3 + 960x^4 - 768x^5 + 256x^6 - x + 12x^2 - 60x^3 + 160x^4 - 240x^5 + 192x^6 - 64x^7 \]
  3. Step 3 — Omit the higher terms and state the result

    Omit the terms containing \(x^3\) and higher since \(x\) is too small so that terms of \(x^3\) and higher can be ignored:

    \[ (4-x)(1-2x)^6 = 4 - 49x + 252x^2 \]

    That is, with terms of \(x^3\) and higher ignored, \((4-x)(1-2x)^6\) is \(4 - 49x + 252x^2\): 4 minus 49 \(x\) plus 252 \(x\) squared.

Writing down an expansion of a series

Question

Write down the expansion of \(\left(3x - \dfrac{1}{3}y\right)^{4}\)

Solution
  1. Step 1 — Set up

    Use the Binomial Theorem to find the expansion of \(\left(3x - \frac{1}{3}y\right)^{4}\). There will be 5 terms in the expansion.

    Use the expansion of \((a+b)^n\) where \(a = 3x\), \(b = -\frac{1}{3}y\) and \(n = 4\).

  2. Step 2 — Write out the expansion
    \[ \left(3x - \frac{1}{3}y\right)^{4} = \binom{4}{0}(3x)^4 + \binom{4}{1}(3x)^3\left(-\frac{1}{3}y\right) + \binom{4}{2}(3x)^2\left(-\frac{1}{3}y\right)^{2} + \binom{4}{3}(3x)\left(-\frac{1}{3}y\right)^{3} + \binom{4}{4}\left(-\frac{1}{3}y\right)^{4} \]
  3. Step 3 — Evaluate the combinations and the indices
    \[ = (1)(81x^4) + (4)(27x^3)\left(-\frac{1}{3}y\right) + (6)(9x^2)\left(\frac{1}{9}y^2\right) + (4)(3x)\left(-\frac{1}{27}y^3\right) + (1)\left(\frac{1}{81}y^4\right) \]
  4. Step 4 — Evaluate the product
    \[ \left(3x - \frac{1}{3}y\right)^{4} = 81x^4 - 36x^3y + 6x^2y^2 - \frac{4}{9}xy^3 + \frac{1}{81}y^4 \]

    That is, the expansion is \(81x^4 - 36x^3y + 6x^2y^2 - \frac{4}{9}xy^3 + \frac{1}{81}y^4\): 81 \(x\) to the power of 4, minus 36 \(x\) cubed \(y\), plus 6 \(x\) squared \(y\) squared, minus four ninths \(x\) \(y\) cubed, plus one over 81 \(y\) to the power of 4.

Finding a specific coefficient in an expansion

Question

Find the coefficient of the term \(x^4\) of the expansion \((7x - 3)^5\)

Solution
  1. Step 1 — Set up

    Use the Binomial Theorem to find the term that will give \(x^4\) in the expansion of \((7x - 3)^5\). We only want to find the coefficient of the term in \(x^4\) so we don't need the complete expansion.

    Let \(a = 7x\), \(b = -3\), \(n = 5\), \(n - k = 4\) (since the index of \(x\) in the 1st term is \(1\)):

  2. Step 2 — To solve for \(k\)

    Substitute \(n = 5\) into \(n - k = 4\):

    \[ 5 - k = 4 \]

    Subtract 5 from both sides:

    \[ -k = 4 - 5 \]

    Subtract 5 from 4 then multiply both sides by \(-1\):

    \[ k = 1 \]
  3. Step 3 — Substitute into the general term

    Substitute \(a\), \(b\), \(n\), \(k\), \(n - k\) into \(\binom{n}{k}a^{n-k}b^k\) to get the term with \(x^4\):

    \[ \binom{n}{k}a^{n-k}b^k = \binom{5}{1}(7x)^4(-3) \]

    Evaluate \(\binom{5}{1}\) and \((7x)^4\):

    \[ = (5)(2401x^4)(-3) \]

    Evaluate the product:

    \[ = -36015x^4 \]
  4. Step 4 — State the coefficient

    Therefore, the coefficient of the term in \(x^4\) in the expansion is \(-36015\): minus 36015.

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