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:

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

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.

\(\\[1pt]\)
\({\small 8.\hspace{0.4em}(\textrm{i}).\hspace{0.5em}}\) Find the first 3 terms in the expansion of \({\small {(2x \ – \ {\large \frac{1}{16x}})}^{8} }\) in descending powers of x.
\(\\[1pt]\)
\({\small\hspace{1em}(\textrm{ii}).\hspace{0.3em}}\) Hence find the coefficient of \({\small {x}^{4}}\) in the expansion of \({\small {(2x \ – \ {\large \frac{1}{16x}})}^{8} {( {\large \frac{1}{{x}^{2}}} \ + \ 1 )}^{2} }\).

\(\\[1pt]\)
\({\small 9.\enspace}\) 9709/31/M/J/20 – Paper 31 June 2020 Pure Maths 3 No 2(a) and (b)
\(\\[1pt]\)
\({\small \hspace{1em}(\textrm{a}).\hspace{0.5em}}\) Expand \( {(2 − 3x)}^{−2} \) in ascending powers of \(x\), up to and including the term in \(x^2\), simplifying the coefficients.
\(\\[1pt]\)
\({\small\hspace{1em}(\textrm{b}).\hspace{0.5em}}\) State the set of values of \(x\) for which the expansion is valid.

\(\\[1pt]\)
\({\small 10.\enspace}\) 9709/33/M/J/21 – Paper 33 June 2021 Pure Maths 3 No 1
\(\\[1pt]\)
Expand \( {(1 + 3x)}^{\frac{2}{3}} \) in ascending powers of \(x\), up to and including the term in \(x^3\), simplifying the coefficients.

\(\\[1pt]\)
\({\small 11.\enspace}\) 9709/32/M/J/21 – Paper 32 June 2021 Pure Maths 3 No 9(a) and (b)
\(\\[1pt]\)
Let \({\small f(x) \ = \ {\large \frac{ 14 \ – \ 3x \ + \ 2x^{2} }{(2 \ + \ x)( 3 \ + \ x^{2} ) }} }\).
\(\\[1pt]\)
\({\small \hspace{1em}(\textrm{a}).\hspace{0.5em}}\) Express \({\small f(x) }\) in partial fractions.
\(\\[1pt]\)
\({\small \hspace{1em}(\textrm{b}).\hspace{0.5em}}\) Hence obtain the expansion of \(f(x)\) in ascending powers of \(x\), up to and including the term in \(x^2\).

\(\\[1pt]\)
\({\small 12.\enspace}\) 9709/13/O/N/21 – Paper 13 Nov 2021 Pure Maths 1 No 2
\(\\[1pt]\)
\({\small \hspace{1em}(\textrm{a}).\hspace{0.5em}}\) Find the first three terms, in ascending powers of \(x\), in the expansion of \( {(1 \ + \ ax)}^{6}\).
\(\\[1pt]\)
\({\small \hspace{1em}(\textrm{b}).\hspace{0.5em}}\) Given that the coefficient of \(x^2\) in the expansion of \( (1 \ – \ 3x){(1 \ + \ ax)}^{6}\) is \(−3\), find the possible values of the constant \(a\).

\(\\[1pt]\)
\({\small 13.\enspace}\) 9709/12/M/J/21 – Paper 12 June 2021 Pure Maths 1 No 4
\(\\[1pt]\)
The coefficient of \(x\) in the expansion of \( \ {(4x \ + \ {\large\frac{10}{x}} )}^{3} \ \) is \(p\). The coefficient of \( {\large\frac{1}{x} }\) in the expansion of \( \ {(2x \ + \ {\large\frac{k}{{x}^{2}} })}^{5} \ \) is \(q\).
\(\\[1pt]\)
Given that \( \ p \ = \ 6q\), find the possible values of \(k\).

\(\\[1pt]\)
\({\small 14.\enspace}\) 9709/12/O/N/19 – Paper 12 November 2019 Pure Maths 1 No 1
\(\\[1pt]\)
The coefficient of \(x^2\) in the expansion of \( \ (4 \ + \ ax){(1 \ + \ {\large\frac{x}{2}})}^{6} \ \) is \(3\).
\(\\[1pt]\)
Find the value of the constant \(a\).

\(\\[1pt]\)


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 =) .