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 =) .

Partial Fractions - Summary of Forms

Partial Fractions

Partial Fractions

In solving algebra related problems and questions, we may sometimes deal with rational functions. A rational function is basically an algebraic polynomial fraction, in which we have polynomials on both the numerator and denominator.

Partial fractions is one of the simplest and most effective method in solving algebra related problems regarding rational functions.

In partial fractions, we separate the polynomials in our rational function into simpler form of polynomials.

Some of the applications of partial fractions include the solving of integration problems with rational functions, the binomial expansion and also the arithmetic series and sequences.

There are a few basic forms we need to memorize in partial fractions:

1.\(\enspace\) The linear form:
\(\hspace{6em} {\small (ax + b)}\)
Example:

\({\large\frac{3x \ + \ 5}{(x \ + \ 1)(2x \ + \ 7)} \ \equiv \ \frac{A}{(x \ + \ 1)} + \frac{B}{(2x \ + \ 7)} }\)

2.\(\enspace\) The quadratic form of a linear factor:
\(\hspace{6em} {\small (cx \ + \ d)^{2}} \)
Example:

\(\frac{3x + 5}{(x + 1){(2x + 7)}^{2}} \equiv \frac{A}{(x + 1)} + {\small\boxed{\frac{B}{(2x + 7)} + \frac{C}{{(2x + 7)}^{2}}}} \)

3.\(\enspace\) The quadratic form that cannot be factorized:
\(\hspace{6em} {\small (c{x}^{2} \ + \ d) }\)
Example:

\({\large\frac{3x \ + \ 5}{(x \ + \ 1)(2{x}^{2} \ + \ 7)} \ \equiv \ \frac{A}{(x \ + \ 1)} + {\small\boxed{\frac{Bx \ + \ C}{(2{x}^{2} \ + \ 7)}}}}\)

After the polynomials in the denominator of the rational function is separated, make the denominators of the simpler terms to be the same. This is typically done by multiplying the denominators together .

To find each of the coefficients in the numerator (A, B or C), we can use substitution method or equating the coefficient method.

In the substitution method, we substitute a value of x that we freely choose in the left hand side numerator and the right hand side numerator and then find the coefficients one by one.

While in the equating the coefficient method, we expand the right hand side numerator and then compare each of the coefficients in the right hand side numerator with the left hand side numerator.

Both methods will be shown in the solution of the examples below. Give it a try 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.
\(\\[1pt]\)

\(\\[1pt]\)
\({\small 2.\enspace}\) Let \({\small f(x) \ = \ {\large \frac{ x \ – \ 4{x}^{2} }{(3 \ – \ x)(2 \ + \ {x}^{2}) }} }\)
\(\\[1pt]\)
Express \({\small f(x) }\) in partial fractions.
\(\\[1pt]\)

\(\\[1pt]\)
\({\small 3.\enspace}\) Let \({\small f(x) \ = \ {\large \frac{ 5{x}^{2} \ + \ x \ + \ 27 }{(2x \ + \ 1)( {x}^{2} \ + \ 9 ) }} }\)
\(\\[1pt]\)
Express \({\small f(x) }\) in partial fractions.
\(\\[1pt]\)

\(\\[1pt]\)
\({\small 4.\enspace}\) Let \({\small f(x) \ = \ {\large \frac{ 10 x \ + \ 9 }{(2x \ + \ 1){( 2x \ + \ 3 )}^{2} }} }\)
\(\\[1pt]\)
Express \({\small f(x) }\) in partial fractions.
\(\\[1pt]\)

\(\\[1pt]\)
\({\small 5.\enspace}\) Let \({\small f(x) \ = \ {\large \frac{ 2x(5 \ – \ x) }{(3 \ + \ x){( 1 \ – \ x )}^{2} }} }\)
\(\\[1pt]\)
Express \({\small f(x) }\) in partial fractions.
\(\\[1pt]\)

\(\\[1pt]\)


PRACTICE MORE WITH THESE QUESTIONS BELOW!

\({\small 1.\enspace}\) Express \({\small {\large \frac{7{x}^{2} \ – \ 3x \ + \ 2}{x({x}^{2} \ + \ 1) }} }\) in partial fractions.

\({\small 2. \enspace}\) Let \({\small f(x) \ = \ {\large \frac{ 5{x}^{2} \ + \ x \ + \ 6 }{(3 \ – \ 2x)({x}^{2} \ + \ 4 )}} }\)
Express \({\small f(x) }\) in partial fractions.

\({\small 3. \enspace}\) Express \({\small {\large \frac{2 \ – \ x \ + \ 8{x}^{2}}{(1 \ – \ x)(1 \ + \ 2x)(2 \ + \ x) }} }\) in partial fractions.

\({\small 4. \enspace}\) Let \({\small f(x) \ = \ {\large \frac{ {x}^{2} \ + \ 3x \ + \ 3 }{(x \ + \ 1)(x \ + \ 3 )}} }\)
\({\small\hspace{1.2em}\left(a\right).\hspace{0.8em}}\) Express f(x) in partial fractions.
\({\small\hspace{1.2em}\left(b\right).\hspace{0.8em}}\) Hence show that,
\(\hspace{3em} {\small \displaystyle \int_{0}^{3} f(x) \ \mathrm{d}x = 3 \ – \ \frac{1}{2} \ln 2.}\)

\({\small 5. \enspace}\) Let \({\small f(x) \ = \ {\large \frac{ {x}^{2} \ – \ 8x \ + \ 9 }{(1 \ – \ x){(2 \ – \ x )}^{2}}} }\)
Express \({\small f(x) }\) in partial fractions.

\({\small 6. \enspace}\) Let \({\small f(x) \ = \ {\large \frac{ 2{x}^{2} \ – \ 7x \ – \ 1 }{(x \ – \ 2)({x}^{2} \ + \ 3 )}} }\)
Express \({\small f(x) }\) in partial fractions.

\(\\[12pt]{\small 7. \enspace}\) Express \({\small {\large \frac{ x \ + \ 5 }{(x \ + \ 1)({x}^{2} \ + \ 3 )}} }\) in the form:
\(\hspace{2em} {\large \frac{A}{( x \ + \ 1)} + \frac{Bx \ + \ C}{ ( {x}^{2} \ + \ 3) } } \)

\({\small 8. \enspace}\) Express in partial fractions \({\small {\large \frac{{x}^{4}}{ {x}^{4} \ – \ 1 }} }\)

\({\small 9. \enspace}\) Express in partial fractions \({\small {\large \frac{{x}^{3} \ + \ x \ – \ 1}{ {x}^{2} \ + \ {x}^{4} }} }\)

\(\\[10pt]{\small 10.\enspace}\) Express in partial fractions
\(\hspace{2em}{\small {\large \frac{x \ + \ 5}{ {x}^{3} \ + \ 5{x}^{2} \ + \ 7x \ + \ 3 }} }\)


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