Question

If ${{\left( 3x-1 \right)}^{7}}={{a}_{7}}{{x}^{7}}+{{a}_{6}}{{x}^{6}}+\cdots +{{a}_{1}}x+{{a}_{0}}$, then $a_7+a_6+a_5+a_4+...+a_1+a_0$= is equal to
[a] 0
[b] 1
[c] 128
[d] 64

Answer 1

Hint: Use the binomial theorem to expand ${{\left( 3x-1 \right)}^{7}}$ and hence calculate the value of the coefficients and hence calculate the sum of the coefficients.

Complete step-by-step answer:
Observe that we are asked to find the sum of the coefficients in the expansion of ${{\left( 3x-1 \right)}^{7}}$
We know that ${{\left( x+y \right)}^{n}}{{=}^{n}}{{C}_{0}}{{x}^{n}}{{+}^{n}}{{C}_{1}}{{x}^{n-1}}y+\ldots {{+}^{n}}{{C}_{n-1}}x{{y}^{n-1}}{{+}^{n}}{{C}_{n}}{{y}^{n}}$
Replace x by 3x and y by -1 and put n =7, we get
${{\left( 3x-1 \right)}^{7}}{{=}^{7}}{{C}_{0}}{{\left( 3x \right)}^{7}}{{+}^{7}}{{C}_{1}}{{\left( 3x \right)}^{6}}\left( -1 \right)+\cdots {{+}^{7}}{{C}_{6}}\left( 3x \right){{\left( -1 \right)}^{6}}{{+}^{7}}{{C}_{7}}{{\left( -1 \right)}^{7}}$
Now, we have
$^{7}{{C}_{0}}=1$
We know that $\dfrac{^{n}{{C}_{r}}}{^{n}{{C}_{r-1}}}=\dfrac{n-r+1}{r}$
Put n=7 and r = 1, we get
$\dfrac{^{7}{{C}_{1}}}{^{7}{{C}_{0}}}=\dfrac{7-1+1}{1}=7{{\Rightarrow }^{7}}{{C}_{1}}=7\times 1=7$
Put n = 7 and r = 2, we get
$\dfrac{^{7}{{C}_{2}}}{^{7}{{C}_{1}}}=\dfrac{7-2+1}{2}{{\Rightarrow }^{7}}{{C}_{2}}=\dfrac{6}{2}\times 7=21$
Put n = 7 and r = 3, we get
$\dfrac{^{7}{{C}_{3}}}{^{7}{{C}_{2}}}=\dfrac{7-3+1}{3}{{\Rightarrow }^{7}}{{C}_{3}}=\dfrac{5}{3}\times 21=35$
Put n = 7 and r = 4, we get
$\dfrac{^{7}{{C}_{4}}}{^{7}{{C}_{3}}}=\dfrac{7-4+1}{4}{{\Rightarrow }^{7}}{{C}_{4}}=\dfrac{4}{4}\times 35=35$
Put n = 7 and r = 5, we get
$\dfrac{^{7}{{C}_{5}}}{^{7}{{C}_{4}}}=\dfrac{7-5+1}{5}{{\Rightarrow }^{7}}{{C}_{5}}=\dfrac{3}{5}\times 35=21$
Put n = 7 and r = 6, we get
$\dfrac{^{7}{{C}_{6}}}{^{7}{{C}_{5}}}=\dfrac{7-6+1}{6}{{\Rightarrow }^{7}}{{C}_{6}}=\dfrac{2}{6}\times 21=7$
Put n = 7 and r = 7, we get
$\dfrac{^{7}{{C}_{7}}}{^{7}{{C}_{6}}}=\dfrac{7-7+1}{7}{{\Rightarrow }^{7}}{{C}_{7}}=\dfrac{1}{7}\times 7=1$
Hence, we have
${{\left( 3x-1 \right)}^{7}}={{\left( 3x \right)}^{7}}-7{{\left( 3x \right)}^{6}}+21{{\left( 3x \right)}^{5}}-35{{\left( 3x \right)}^{4}}+35{{\left( 3x \right)}^{3}}-21{{\left( 3x \right)}^{2}}+7\left( 3x \right)-1$
Hence, we have
${{\left( 3x-1 \right)}^{7}}=2187{{x}^{7}}-5103{{x}^{6}}+5103{{x}^{5}}-2835{{x}^{4}}+945{{x}^{3}}-189{{x}^{2}}+21x-1$
Hence, we have
$\sum\limits_{r=0}^{7}{{{a}_{r}}}=2187-5103+5103-2835+945-189+21-1=128$
Hence the sum of the coefficients is equal to 128
Hence option [c] is correct.

Note: Alternative solution: Best method:
We have ${{\left( 3x-1 \right)}^{7}}={{a}_{7}}{{x}^{7}}+{{a}_{6}}{{x}^{6}}+\cdots +{{a}_{1}}x+{{a}_{0}}$
Put x = 1, we get
${{\left( 3\times 1-1 \right)}^{7}}={{a}_{7}}{{\left( 1 \right)}^{7}}+{{a}_{6}}{{\left( 1 \right)}^{6}}+\cdots +{{a}_{1}}\left( 1 \right)+{{a}_{0}}$
Hence, we have
${{a}_{7}}+{{a}_{6}}+\cdots +{{a}_{1}}+{{a}_{0}}={{\left( 2 \right)}^{7}}=128$
Hence the sum of the coefficients is equal to 128.
Hence option [c] is correct.