Question

Explain binomial series .

Answer 1

Hint : We have to state what binomial series is . We also have to give the expression for the expansion of sum or difference of two terms using the binomial theorem . I also use the concept of permutation and combination for expanding the binomial theorem . We also state the concept of Pascal’s Triangle .

Complete step-by-step answer :
Definition of binomial series :
Binomial series is the method of expanding the terms of sum or difference of two variables of an expansion which are raised to any finite power $\left( {{\text{ }}n{\text{ }}} \right),$ where n belongs to natural number . The expansion is used in various sections such as trigonometric formulas , algebra , probability etc .
The expansion of sum of two variables is given by the formula :
${(x + y)^n} = n{C_0}{x^n} + n{C_1}{x^{(n - 1)}} \times y + n{C_2}{x^{(n - 2)}} \times {y^2} + ............... + n{C_n}{y^n}$
The expansion of difference of two variables is given by the formula :
${(x - y)^n} = n{C_0}{x^n} + ( - 1) \times n{C_1}{x^{(n - 1)}} \times y + n{C_2}{x^{(n - 2)}} \times {y^2} + ............. + {( - 1)^n} \times n{C_n}{y^n}$
The expansion for various value of n :
${(A + B)^0} = 1$
${(A + B)^1} = A + B$
${(A + B)^2} = {A^2} + 2AB + {B^2}$
${(A + B)^3} = {A^3} + 3 \times {A^2} \times B + 3 \times A \times {B^2} + {B^3}$
${(A + B)^4} = 4{C_0}{A^4} + 4{C_1}{A^3} \times B + 4{C_2}{A^2} \times {B^2} + 4{C_3}A \times {B^3} + 4{C_4}{B^4}$
( using the values of combinations )
${(A + B)^4} = {A^4} + 4 \times {A^3} \times B + 6{A^2} \times {B^2} + 4A \times {B^3} + {B^4}$
From the above expression we can conclude that the total number of terms in the expression are $1$ more than the power of the expression .
The sum of powers of variables of the expansion should be equal to the power of the original term .
Some other expressions for the expansion :
${(x + y)^n} + {(x + y)^n} = 2 \times [n{C_0}{x^n} + n{C_2}{x^{(n - 1)}}{y^2} + n{C_4}{x^{(n - 4)}}{y^4} + .............]$ ———(1)
${(x + y)^n} - {(x - y)^n} = 2 \times [n{C_1}{x^{(n - 1)}}y + n{C_3}{x^{(n - 3)}}{y^3} + n{C_5}{x^{(n - 5)}}{y^5} + .............]$ ———(2)
${(1 + x)^n} = [n{C_0} + n{C_1}x + n{C_2}{x^2} + {\text{ }}....... + n{C_n}{x^n}]$
${(1 + x)^n} + {(1 - x)^n} = 2 \times [n{C_0} + n{C_2}{x^2} + n{C_4}{x^4} + ..........]$
${(1 + x)^n} - {(1 - x)^n} = 2 \times [n{C_1}x + n{C_3}{x^3} + n{C_5}{x^5} + .................]$
The number of terms in the expansion$\;\left( 1 \right)$are $\dfrac{{\left( {{\text{ }}n{\text{ }} + {\text{ }}2{\text{ }}} \right){\text{ }}}}{2}$if is even or$\dfrac{{\left( {{\text{ }}n{\text{ }} + {\text{ 1 }}} \right){\text{ }}}}{2}$ if is odd .
The number of terms in the expansion $\left( 2 \right)$are $\dfrac{n}{2}$if is even or$\dfrac{{\left( {{\text{ }}n{\text{ }} + {\text{ 1 }}} \right){\text{ }}}}{2}$ if is odd .

Note : Corresponding to each combination of ${}^n{C_r}$we have $r!$permutations, because $r$ objects in every combination can be rearranged in $r!$ways . Hence , the total number of permutations of $n$ different things taken $r$ at a time is${}^n{C_r}{\text{ }} \times {\text{ }}r!$. Thus ${}^n{P_r}{\text{ }} = {\text{ }}{}^n{C_r}{\text{ }} \times {\text{ }}r!{\text{ }},{\text{ }}0 < {\text{ }}r{\text{ }} \leqslant n\;$
Also , some formulas used :
${}^n{C_1}{\text{ }} = {\text{ }}n$
${}^n{C_2} = \dfrac{{{\text{ }}n\left( {n - 1} \right)}}{2}$
${}^n{C_0}{\text{ }} = {\text{ }}1$
${}^n{C_n} = {\text{ }}1$