Question

If in the expansion of \[{{\left( {{3}^{\dfrac{-x}{4}}}+{{3}^{\dfrac{5x}{4}}} \right)}^{n}}\]the sum of binomial Coefficient is 64 then value of n is
a) 6
b) 7
c) 8
d) 9

Answer 1

Hint: first write the general term of expansion and the then find the expression for some of the binomial coefficient then find the expansion of ${{\left( 1+x \right)}^{n}}$ by putting the value of $x=1$ Compare the equation to get the answer.

Complete step by step solution: First of all we will write general term of expansion of \[{{\left( {{3}^{\dfrac{-x}{4}}}+{{3}^{\dfrac{5x}{4}}} \right)}^{n}}\]
\[T_{r+1}={}^{n}{{C}_{r}}{{\left( {{3}^{\dfrac{-x}{4}}} \right)}^{n-r}}{{\left( {{3}^{\dfrac{5x}{4}}} \right)}^{r}}\]
Here binomial Coefficient is ${}^{n}{{C}_{r}}$.
So it is given that
$\sum\limits_{r=0}^{n}{{}^{n}{{C}_{r}}}=64$
Expanding
\[{}^{n}{{C}_{0}}+{}^{n}{{C}_{1}}+{}^{n}{{C}_{2}}+\_\ \_\ \_+{}^{n}{{C}_{n}}=64\] ----(1)
Now we know that the Binomial expansion of ${{(1+x)}^{n}}$.
${{\left( 1+x \right)}^{n}}=\sum\limits_{r=0}^{n}{{}^{n}{{C}_{r}}}{{\left( 1 \right)}^{n-r}}{{x}^{r}}$
\[{{\left( 1+x \right)}^{n}}={}^{n}{{C}_{0}}{{\left( 1 \right)}^{n}}\ \centerdot \ {{x}^{0}}+{}^{n}{{C}_{1}}{{\left( 1 \right)}^{n-1}}\ \centerdot \ {{x}^{1}}+\_\ \_\ \_+{}^{n}{{C}_{n}}{{\left( 1 \right)}^{n-n}}\ \centerdot \ {{x}^{n}}\]
  \[{{\left( 1+x \right)}^{n}}={}^{n}{{C}_{0}}\ +{}^{n}{{C}_{1}}\ \left( x \right)+{}^{n}{{C}_{2}}{{\left( x \right)}^{2}}\_\ \_\ \_+{}^{n}{{C}_{n}}\ {{\left( x \right)}^{n}}\]
Now putting $x=1$ we get
\[\Rightarrow {{\left( 1+1 \right)}^{n}}={}^{n}{{C}_{0}}+{}^{n}{{C}_{1}}\,\centerdot \ 1+{}^{n}{{C}_{2}}\centerdot \ {{\left( 1 \right)}^{2}}+\_\ \_\ \_+{}^{n}{{C}_{n}}{{\left( 1 \right)}^{n}}\]
\[\Rightarrow {{2}^{n}}={}^{n}{{C}_{0}}+{}^{n}{{C}_{1}}\,+{}^{n}{{C}_{2}}+\_\ \_\ \_+{}^{n}{{C}_{n}}\]-------(2)
We get (1) = (2)
So, \[{}^{n}{{C}_{0}}+{}^{n}{{C}_{1}}\,+{}^{n}{{C}_{2}}+\_\ \_\ \_+{}^{n}{{C}_{n}}=64={{2}^{n}}\]
\[\Rightarrow {{2}^{n}}=64\]
Since ${{2}^{6}}=64$
So, $\Rightarrow {{2}^{n}}={{2}^{\left( 6 \right)}}$
Comparing LHS and RHS
We get n=6
So n =6 is the answer.

Note: the general formula for expansion is
${{\left( x+y \right)}^{n}}=\sum\limits_{r=0}^{n}{{}^{n}{{C}_{r}}}{{x}^{n-r}}{{y}^{r}}$ where
${}^{n}{{C}_{r}}=\dfrac{n!}{\left( n-r \right)!r!}$
Here $n\in N,\ x,\ y\in R$.