Question

How do you use the binomial series to expand ${(2 + x)^{11}}$ ?

Answer 1

Hint: Binomial is a polynomial expression having two terms. We have to expand the given binomial expression raised to the power of $11$. For this purpose we have to use the binomial series expansion formula also known as Binomial Theorem. For $a$ and $b$ given as any term and $n$ given as any positive integer, Binomial theorem states,
\[{\left( {a + b} \right)^n} = \sum\nolimits_{r = 0}^n {{}^n{C_r}{a^{n - r}}{b^r}} \]

Formula Used:
\[
  {\left( {a + b} \right)^n} = \sum\nolimits_{r = 0}^n {{}^n{C_r}{a^{n - r}}{b^r}} \\
  {}^n{C_r} = \dfrac{{n!}}{{\left( {n - r} \right)!\; \times r!}} \\
 \]

Complete step by step solution:
We have been given an expression ${(2 + x)^{11}}$ which is a binomial expression $(2 + x)$raised to the power $11$. We have to expand this expression using binomial series (or binomial theorem).
The general formula for expansion of a binomial expression raised to some positive integer power is given as,
\[{\left( {a + b} \right)^n} = \sum\nolimits_{r = 0}^n {{}^n{C_r}{a^{n - r}}{b^r}} \]
The expanded series will contain $n + 1$ terms and any ${r^{th}}$ term of the expanded series will be of the form \[{}^n{C_r}{a^{n - r}}{b^r}\].
On comparing with the given expression we can observe that,
$a = 2$, $b = x$ and $n = 11$.
Using the general form of binomial expansion formula we can write the given expression as,
\[{\left( {2 + x} \right)^{11}} = \sum\nolimits_{r = 0}^{11} {{}^{11}{C_r}{2^{11 - r}}{x^r}} \]
Now we have to expand the summation. The value of $r$ varies from $0$ to $11$, and correspondingly each term will vary. We can expand the summation as,
\[
  \sum\nolimits_{r = 0}^{11} {{}^{11}{C_r}{2^{n - r}}{x^r}} \\
   = {}^{11}{C_0}{2^{11 - 0}}{x^0} + {}^{11}{C_1}{2^{11 - 1}}{x^1} + {}^{11}{C_2}{2^{11 - 2}}{x^2} + {}^{11}{C_3}{2^{11 - 3}}{x^3} \\
   + {}^{11}{C_4}{2^{11 - 4}}{x^4} + {}^{11}{C_5}{2^{11 - 5}}{x^5} + {}^{11}{C_6}{2^{11 - 6}}{x^6} + {}^{11}{C_7}{2^{11 - 7}}{x^7} \\
   + {}^{11}{C_8}{2^{11 - 8}}{x^8} + {}^{11}{C_9}{2^{11 - 9}}{x^9} + {}^{11}{C_{10}}{2^{11 - 10}}{x^{10}} + {}^{11}{C_{11}}{2^{11 - 11}}{x^{11}} \\
 \]
We have expanded the given expression using the binomial series expansion.
Now we try to simplify each term. We will use the formula \[{}^n{C_r} = \dfrac{{n!}}{{\left( {n - r} \right)!\; \times r!}}\].
The expanded series in simplified form becomes,
\[
   = \dfrac{{11!}}{{11!\; \times 0!}}{2^{11}} + \dfrac{{11!}}{{10!\; \times 1!}}{2^{10}}x + \dfrac{{11!}}{{9!\; \times 2!}}{2^9}{x^2} + \dfrac{{11!}}{{8!\; \times 3!}}{2^8}{x^3} \\
   + \dfrac{{11!}}{{7!\; \times 4!}}{2^7}{x^4} + \dfrac{{11!}}{{6!\; \times 5!}}{2^6}{x^5} + \dfrac{{11!}}{{5!\; \times 6!}}{2^5}{x^6} + \dfrac{{11!}}{{4!\; \times 7!}}{2^4}{x^7} \\
   + \dfrac{{11!}}{{3!\; \times 8!}}{2^3}{x^8} + \dfrac{{11!}}{{2!\; \times 9!}}{2^2}{x^9} + \dfrac{{11!}}{{1!\; \times 10!}}{2^1}{x^{10}} + \dfrac{{11!}}{{0!\; \times 11!}}{2^0}{x^{11}} \\
   = {2^{11}} + 11 \times {2^{10}}x + 55 \times {2^9}{x^2} + 165 \times {2^8}{x^3} + 330 \times {2^7}{x^4} + 462 \times {2^6}{x^5} + 462 \times {2^5}{x^6} + 330 \times {2^4}{x^7} \\
   + 165 \times {2^3}{x^8} + 55 \times {2^2}{x^9} + 11 \times {2^1}{x^{10}} + {x^{11}} \\
   = 2048 + 11264x + 28160{x^2} + 42240{x^3} + 42240{x^4} + 29568{x^5} + 14784{x^6} + 5280{x^7} + 1320{x^8} \\
   + 220{x^9} + 22{x^{10}} + {x^{11}} \\
 \]
Thus, we get the expanded series as,
\[
  {(2 + x)^{11}} = 2048 + 11264x + 28160{x^2} + 42240{x^3} + 42240{x^4} + 29568{x^5} + 14784{x^6} + 5280{x^7} + 1320{x^8} \\
   + 220{x^9} + 22{x^{10}} + {x^{11}} \\
 \]
Note: We used the formula for binomial series expansion to find the expanded series of the given expression. The expanded series contains $n + 1$ terms. For negative powers we have to use different formulas for expansion. If not specified in the question we may not have to calculate the exact coefficients of each term as it may be rigorous to multiply such huge numbers.