Binomial expansion formula

Binomial Expansion and Binomial Series

Binomial Expansion and Binomial Series

In algebra, we all have learnt the following basic algebraic expansion:

\( \hspace{3em} {(a + b)}^{2} = {a}^{2} + 2ab + {b}^{2} \).

We can keep multiplying the expression \( { \small (a + b) } \) by itself to find the expression for higher index value. For example:

\(\\[12pt] {(a + b)}^{3} \ = \ {(a + b)}^{2} \times (a + b) \)
\(\\[12pt] \hspace{1.5em} \ = \ ({a}^{2} + 2ab + {b}^{2}) \times (a + b) \)
\(\\[12pt] \hspace{1.5em} \ = \ {a}^{3} + {a}^{2}b + 2{a}^2b + 2a{b}^2 + a{b}^2 + {b}^{3} \)
\( \hspace{1.5em} \ = \ {a}^{3} + 3{a}^{2}b + 3a{b}^2 + {b}^{3} \)

Of course this is a really tedious way for a very large index/power/exponent number.

Instead of doing it manually, we could use a formula called the Binomial Theorem which is shown below:

\( {(a + b)}^{n} \ = \ \displaystyle \sum_{k=0}^{n} \binom{n}{k} {a}^{n \ – \ k} \ {b}^{k} \)

with \( \displaystyle \binom{n}{k} \ = \ \frac{n!}{k!(n-k)!} \)

\(\hspace{4.4em} = \ {\large \frac{n (n \ – \ 1) (n \ – \ 2) … (n \ – \ k \ + \ 1)}{k(k \ – \ 1)(k \ – \ 2) \ … \ 1} } \)

for \( k \geq 1 \) and \( \displaystyle \binom{n}{0} \ = \ 1 \)

\(\\[10pt]\) Example:
\(\\[12pt] {(a + b)}^{10} \)
\(\\[15pt] = \binom{10}{0} {a}^{10}{b}^{0} + \binom{10}{1} {a}^{9}{b}^{1}+…+ \binom{10}{10} {a}^{0}{b}^{10} \)
\(\\[15pt] = {a}^{10} + 10 {a}^{9}b+ 45 {a}^{8}{b}^{2} + … + 10{a}{b}^{9} + {b}^{10} \)

\(\\[10pt]\) Note that,
\(\\[20pt]\quad \ {\displaystyle \binom{10}{0} } \ = \ 1 \)
\(\\[20pt]\quad \ {\displaystyle \binom{10}{1} } \ = \ {\large \frac{ 10 }{ 1 } } \ = \ 10 \)
\(\\[20pt]\quad \ {\displaystyle \binom{10}{2} } \ = \ {\large \frac{ 10 \ \times \ 9 }{ 2 \ \times \ 1 } } \ = \ 45 \)
\(\\[20pt]\quad \ {\displaystyle \binom{10}{3} } \ = \ {\large \frac{ 10 \ \times \ 9 \ \times \ 8 }{ 3 \ \times \ 2 \ \times \ 1 } } \ = \ 120 \)
\(\\[20pt]\quad \ {\displaystyle \binom{10}{4} } \ = \ {\large \frac{ 10 \ \times \ 9 \ \times \ 8 \ \times \ 7 }{4 \ \times \ 3 \ \times \ 2 \ \times \ 1 } } \ = \ 210 \)
\(\\[20pt]\quad \ {\displaystyle \binom{10}{5} } \ = \ {\large \frac{10 \ \times \ 9 \ \times \ 8 \ \times \ 7 \ \times \ 6 }{5 \ \times \ 4 \ \times \ 3 \ \times \ 2 \ \times \ 1 } } \ = \ 252 \)
, etc.

\(\binom{n}{k} \) is read as “n choose k” or sometimes referred to as the binomial coefficients.

Notice that this binomial expansion has a finite number of terms with the k values take the non-negative numbers from 0, 1, 2, … , n.

Then the next question would be: Can we still use the binomial theorem for the expansion with negative number or fractional number for the index value?

Thankfully, Sir Isaac Newton has shown that the binomial theorem can be generalized to take in any numbers for the index value including the negative and fractional numbers as long as it is within a convergence rule.

Using this result, we have the Binomial Series which can be expressed as follows:

\(\\[25pt]{(1 + x)}^{n} = \displaystyle \sum_{k=0}^{n} \binom{n}{k} {x}^{k} \)
\(\\[10pt]{\small = \ 1 + nx + \frac{(n)(n-1)}{2!}{x}^{2} + \frac{(n)(n-1)(n-2)}{3!}{x}^{3} + … }\)

with \( {\small k } \) is any real number and \( {\small -1 \lt x \lt 1 } \).

\(\\[10pt]\) Example:
\(\\[12pt] { {\large \frac{1}{1 + x}} \ = \ (1 + x)}^{-1} \)
\(\\[15pt]{\small = \ 1 + (-1)x + \frac{(-1)(-2)}{2!}{x}^{2} + \frac{(-1)(-2)(-3)}{3!}{x}^{3} + … }\)
\(\\[15pt]{\small = \ 1 \ – \ x \ + \ {x}^{2} \ – \ {x}^{3} \ + \ … \ – \ … }\)
\(\\[15pt]\) The expansion is valid when \({\small -1 \lt x \lt 1 } \).

\(\\[10pt]\) Another example:
\(\\[12pt] { \frac{1}{\sqrt{(1 + x)}} \ = \ (1 + x)}^{\frac{1}{2}} \)
\(\\[15pt]{\small = \ 1 + \frac{1}{2}x + \frac{(\frac{1}{2})(- \frac{1}{2})}{2!}{x}^{2} + \frac{(\frac{1}{2})(- \frac{1}{2})( – \frac{3}{2})}{3!}{x}^{3} + … }\)
\(\\[15pt]{ = \ 1 + \frac{x}{2} – \frac{{x}^{2}}{8} + \frac{{x}^{3}}{16} \ – \ … + … }\)
\(\\[15pt]\) The expansion is valid when \({\small -1 \lt x \lt 1 } \).

Try some of the examples below and if you need any help, just look at the solution I have written. Cheers ! =) .
\(\\[1pt]\)


EXAMPLE:

\({\small 1.\enspace}\) Let \({\small f(x) \ = \ {\large \frac{7{x}^{2} \ – \ 15x \ + \ 8}{(1 \ – \ 2x){(2 \ – \ x)}^{2} }} }\)
\(\\[1pt]\)
Express \({\small f(x) }\) in partial fractions and hence obtain the expansion of \({\small f(x) }\) in ascending powers of x, up to and including the term in \({\small {x}^{2}}\).

\(\\[1pt]\)
\({\small 2.\enspace}\) Let \({\small f(x) \ = \ {\large \frac{ x \ – \ 4{x}^{2} }{(3 \ – \ x)(2 \ + \ {x}^{2}) }} }\)
\(\\[1pt]\)
Express \({\small f(x) }\) in partial fractions and hence obtain the expansion of \({\small f(x) }\) in ascending powers of x, up to and including the term in \({\small {x}^{3}}\).

\(\\[1pt]\)
\({\small 3.\enspace}\) Find the value of \( {\small {(2 \ + \ x)}^{6} \ – \ {(2 \ – \ x)}^{6} }\) in ascending powers of x, up to and including the term in \({\small {x}^{3}}\). Hence, find the value of \( {\small {(1.99)}^{6} \ – \ {(2.01)}^{6} }\).

\(\\[1pt]\)
\({\small 4.\enspace}\) Find the last four terms in the expansion in ascending powers of x of \( (2x – {x}^{2})^{10}\).

\(\\[1pt]\)
\({\small 5.\enspace}\) Find the value of \( {\small \frac{a}{b} }\) in \({\small {(a + bx)}^{12}}\) given that the coefficient of \({\small {x}^{2} }\) is 11 times the coefficient of \({\small x }\).

\(\\[1pt]\)
\({\small 6.\enspace}\) Find the value of a, b and n in the following expansion:
\(\\[1pt]\)
\({\small {(a + bx)}^{n} \ = \ … + {\large\frac{20}{3}}{x}^{2} + 20 {x}^{3} + … }\).
\(\\[1pt]\)
given that \( {\small {\large \frac{a}{b}} \ = \ {\large \frac{4}{9}} }\).

\(\\[1pt]\)
\({\small 7.\enspace}\) Consider the binomial expansion of \({\small {\large \frac{1}{\sqrt{4-x}} } }\).
\(\\[1pt]\)
\({\small\hspace{1.2em}(\textrm{a}).\hspace{0.8em}}\) Write down the first 4 terms.
\({\small\hspace{1.2em}(\textrm{b}).\hspace{0.8em}}\) State the interval of convergence for the complete expression.
\({\small\hspace{1.2em}(\textrm{c}).\hspace{0.8em}}\) Use the expansion to estimate \({\small {\large \frac{1}{\sqrt{3.6}} } }\). Check your answer by direct calculation.

\(\\[1pt]\)
\({\small 8.\hspace{0.4em}(\textrm{i}).\hspace{0.4em}}\) Find the first 3 terms in the expansion of \({\small {(2x \ – \ {\large \frac{1}{16x}})}^{8} }\) in descending powers of x.
\(\\[1pt]\)
\({\small\hspace{1em}(\textrm{ii}).\hspace{0.7em}}\) Hence find the coefficient of \({\small {x}^{4}}\) in the expansion of \({\small {(2x \ – \ {\large \frac{1}{16x}})}^{8} {( {\large \frac{1}{{x}^{2}}} \ + \ 1 )}^{2} }\).

\(\\[1pt]\)


PRACTICE MORE WITH THESE QUESTIONS BELOW!

\({\small 1.\enspace}\) Expand \(\frac{1}{\sqrt[3]{(1 \ + \ 6x)}}\) in ascending powers of x up to and including the term in \({\small x^3 }\), simplifying the coefficients.

\({\small 2. \enspace}\) Expand \(\frac{4}{\sqrt{(4 \ – \ 3x)}}\) in ascending powers of x up to and including the term in \({\small x^2 }\), simplifying the coefficients.

\({\small 3. \enspace}\) Let \({\small f(x) \ = \ {\large \frac{3{x}^{2} \ + \ x \ + \ 6}{(x \ + \ 2)({x}^{2} \ + \ 4) }} }\)

(i). Express \({\small f(x) }\) in partial fractions.
(ii). Hence obtain the expansion of \({\small f(x) }\) in ascending powers of x, up to and including the term in \({\small {x}^{2}}\).

\({\small 4. \enspace}\) Express \( {\small {(3 \ + \ 4x)}^{\frac{1}{2}} }\) as a series of descending powers of x up to and including the third non-zero coefficient. State the set of values of x for which the series expansion is valid.

\({\small 5. \enspace}\) (i). Given that the first three terms in the expansion of \( {\small {(1 \ + \ ax)}^{b}}\) in ascending powers of x are \({\small 1 \ + \ x \ + \ \frac{3}{2}{x}^{2}}\), where a and b are constants, find the values of a and b.
(ii). Hence, with the values of a and b found in (i), expand \(\frac{ {(1 \ + \ ax)}^{b} }{1 \ – \ x}\) as a series in ascending powers of x up to and including the term in \({\small x^2 }\).

\({\small 6. \enspace}\) Expand \({\small {(1-2p)}^{8} }\) up to and including the term in \({\small p^3 }\). Hence, find the first four terms in the expansion in ascending powers of x of \({\small {(1 \ – \ 4x \ + \ \frac{2}{x})}^{8} }\).

\({\small 7. \enspace}\) Find the coefficient of \({\small {x}^{2}}\) in the expansion of \({\small {(1 \ + \ 3x)}^{2}{(1 \ – \ 3x)}^{10} }\).

\({\small 8. \enspace}\) Find the coefficient of \({ \small x }\) in the expansion of \({\small {(3x \ – \ \frac{2}{x})}^{9} }\).

\({\small 9. \enspace}\) Find the value of a and b in the following expansion:

\({\small \hspace{1.5em} {(a + bx)}^{9} \ = \ … + 672{x}^{3} + 252 {x}^{4} + … }\)

\({\small 10.\enspace}\) Find the value of a and b in the expansion of \({\small {(1 \ + \ ax)}^{5}{(1 \ – \ bx)}^{7} }\) if the coefficients of \({ \small x }\) and \({\small {x}^{2}}\) are 3 and -9 respectively.


As always, if you have any particular questions to discuss, leave it in the comment section below. Cheers =) .

8 comments
  1. 1/3√1-x

  2. Can you please expand this one ASAP to the power of x^3 . 3(1-x)^-1 – ((6x+1)(2x^2 – 1)^-1)

  3. 11+x = (1+x)−1
    = 1+((−1)x+(−1)(−2))/2!x^2+((−1)(−2)(−3))/3!x3+…
    = 1 – x + x2 – x3 + … – …
    this example question has a problem or maybe I am wrong, but when the square root goes to the nominator then doesn’t it become (-1/2) and not (1/2)

    1. Hi Ifrah

      There is no square root in the binomial expansion or maybe if you could clarify further what was the question again?

      Cheers,
      Mr Will

  4. where can we get the answers for the questions at the bottom

    1. Hi, haven’t particularly made one
      but we can discuss it here tho
      Will

      1. do you have the answer for question 2

  5. the binomial of (2 –x) 6 in ascending powers of x

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