Question

Prove that the coefficient of $ {x^5} $ in the expansion of $ {(1 + {x^2})^5} \cdot {(1 + x)^4} $ is $ 60. $

Answer 1

Hint: The binomial expansion or the binomial theorem describes the algebraic expansion of the powers of the binomial (binomial is the pair of two terms). Use formula $ {(a + b)^n} = {}^n{C_a}{a^n} + {}^n{C_1}{a^{n - 1}}{b^1} + ..... $ for binomial expansion. Where, $ {}^n{C_a} $ represents the total number of possible ways and use of the laws of powers and exponent accordingly.

Complete step-by-step answer:
By using the formula of the binomial expansion –
 $ {(a + b)^n} = {}^n{C_a}{a^n} + {}^n{C_1}{a^{n - 1}}{b^1} + ..... $
Now, take given binomial expansion and apply the above formula in it –
 $
  {(1 + {x^2})^5} \cdot {(1 + x)^4} = \\
  ({}^5{C_0}(1) + {}^5{C_1}(1){x^2} + {}^5{C_2}(1){\left( {{x^2}} \right)^2} + {}^5{C_3}(1){\left( {{x^2}} \right)^3} + {}^5{C_4}(1){\left( {{x^2}} \right)^4} + {}^5{C_0}(1){\left( {{x^2}} \right)^5}) \times \\
  ({}^4{C_0}(1) + {}^4{C_1}(x) + {}^4{C_2}{(x)^2} + {}^4{C_3}{(x)^3} + {}^4{C_4}{(x)^4}) \\
  $
Simplify the above equation –
 \[
  {(1 + {x^2})^5} \cdot {(1 + x)^4} = \\
  ({}^5{C_0}(1) + \underline{\underline {{}^5{C_1}{x^2}}} + \underline {{}^5{C_2}({x^4})} + {}^5{C_3}({x^6}) + {}^5{C_4}({x^8}) + {}^5{C_0}({x^{10}})) \times \\
  ({}^4{C_0}(1) + \underline {{}^4{C_1}(x)} + {}^4{C_2}({x^2}) + \underline{\underline {{}^4{C_3}({x^3})}} + {}^4{C_4}({x^4})) \\
 \]
Separate the terms which makes the coefficient of $ {x^5} $ . Make the pair of terms which can make the required term.
 \[\{ \underline{\underline {{}^5{C_1}{x^2}}} + \underline {{}^5{C_2}({x^4})\} } . \{ \underline {{}^4{C_1}(x)} + \underline{\underline {{}^4{C_3}({x^3})}} \} \]
Pair of terms which can make the required terms –
 \[\underline {{}^5{C_1}{x^2} \times {}^4{C_3}{x^3}} + \underline {{}^5{C_2}{x^4} \times {}^4{C_1}{x^1}} \]
Exponents with the same base and in the multiplication, then powers are added.
 $ \Rightarrow \left( {{}^5{C_1} \times {}^4{C_3}} \right){x^{2 + 3}} + \left( {{}^5{C_2} \times {}^4{C_1}} \right){x^{1 + 4}} $
Again simplification –
 $ \Rightarrow \left( {5 \times 4} \right){x^5} + \left( {\dfrac{{5 \times 4}}{{2 \times 1}} \times 4} \right){x^5} $
Multiply the constant terms and simplify –
 $ \Rightarrow \left( {20} \right){x^5} + \left( {40} \right){x^5} $
Take variable constant –
 \[
   \Rightarrow {x^5}\left( {20 + 40} \right) \\
   \Rightarrow {x^5}(60) \\
   \Rightarrow 60{x^5} \;
 \]
Hence, the coefficient of $ {x^5} $ is $ 60 $
So, the correct answer is “60”.

Note: Remember the laws of power and exponents, factorial properly and use it wisely. Be careful while opening the brackets and observe each and every term which can make the required term $ {x^5} $ . The segregation of the terms on the basis of the power and exponent is the most important step.
 Know the difference between the permutations and combinations and apply its formula accordingly. In permutations, specific order and arrangement is the most important whereas a combination is used if the certain objects are to be arranged in such a way that the order of objects is not important.
Formula for combinations - $ ^nc{}_r = \dfrac{{n!}}{{r!(n - r)!}} $
Formula for the permutations - $ {}^np{}_r = \dfrac{{n!}}{{(n - r)!}} $