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Question: If in the expansion of \[{{\left( {{3}^{\dfrac{-x}{4}}}+{{3}^{\dfrac{5x}{4}}} \right)}^{n}}\]the sum...

If in the expansion of (3−x4+35x4)n{{\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

Explanation

Solution

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

Complete step by step solution: First of all we will write general term of expansion of (3−x4+35x4)n{{\left( {{3}^{\dfrac{-x}{4}}}+{{3}^{\dfrac{5x}{4}}} \right)}^{n}}
Tr+1=nCr(3−x4)n−r(35x4)rT_{r+1}={}^{n}{{C}_{r}}{{\left( {{3}^{\dfrac{-x}{4}}} \right)}^{n-r}}{{\left( {{3}^{\dfrac{5x}{4}}} \right)}^{r}}
Here binomial Coefficient is nCr{}^{n}{{C}_{r}}.
So it is given that
∑r=0nnCr=64\sum\limits_{r=0}^{n}{{}^{n}{{C}_{r}}}=64
Expanding
nC0+nC1+nC2+  +nCn=64{}^{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{{(1+x)}^{n}}.
(1+x)n=∑r=0nnCr(1)n−rxr{{\left( 1+x \right)}^{n}}=\sum\limits_{r=0}^{n}{{}^{n}{{C}_{r}}}{{\left( 1 \right)}^{n-r}}{{x}^{r}}
(1+x)n=nC0(1)n ⋅ x0+nC1(1)n−1 ⋅ x1+  +nCn(1)n−n ⋅ xn{{\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}}
(1+x)n=nC0 +nC1 (x)+nC2(x)2  +nCn (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=1x=1 we get
⇒(1+1)n=nC0+nC1 ⋅ 1+nC2⋅ (1)2+  +nCn(1)n\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}}
⇒2n=nC0+nC1 +nC2+  +nCn\Rightarrow {{2}^{n}}={}^{n}{{C}_{0}}+{}^{n}{{C}_{1}}\,+{}^{n}{{C}_{2}}+\\_\ \\_\ \\_+{}^{n}{{C}_{n}}-------(2)
We get (1) = (2)
So, nC0+nC1 +nC2+  +nCn=64=2n{}^{n}{{C}_{0}}+{}^{n}{{C}_{1}}\,+{}^{n}{{C}_{2}}+\\_\ \\_\ \\_+{}^{n}{{C}_{n}}=64={{2}^{n}}
⇒2n=64\Rightarrow {{2}^{n}}=64
Since 26=64{{2}^{6}}=64
So, ⇒2n=2(6)\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
(x+y)n=∑r=0nnCrxn−ryr{{\left( x+y \right)}^{n}}=\sum\limits_{r=0}^{n}{{}^{n}{{C}_{r}}}{{x}^{n-r}}{{y}^{r}} where
nCr=n!(n−r)!r!{}^{n}{{C}_{r}}=\dfrac{n!}{\left( n-r \right)!r!}
Here n∈N, x, y∈Rn\in N,\ x,\ y\in R.