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Question: The coefficient of \({{x}^{6}}\) in \[\left[ {{\left( 1+x \right)}^{6}}+{{\left( 1+x \right)}^{7}}+....

The coefficient of x6{{x}^{6}} in [(1+x)6+(1+x)7+....(1+x)15]\left[ {{\left( 1+x \right)}^{6}}+{{\left( 1+x \right)}^{7}}+....{{\left( 1+x \right)}^{15}} \right] is

& A.{}^{16}{{C}_{9}} \\\ & B.{}^{16}{{C}_{5}}-{}^{6}{{C}_{5}} \\\ & C.{}^{16}{{C}_{6}}-1 \\\ & D.~\text{None of these} \\\ \end{aligned}$$
Explanation

Solution

At first use the formula of sum of geometric progression which is a (rn - 1)(r - 1)\dfrac{a\,({{r}^{n}}\text{ - 1)}}{(r\text{ - 1)}}where a is 1st{{\text{1}}^{\text{st}}} term, r is common ratio and n is number of term. Then use the formula for general term of the binomial expansion which is Tr+1 = nCr an−r br{{T}_{r+1}}\text{ = }{}^{n}{{C}_{r}}\text{ }{{\text{a}}^{n-r}}\text{ }{{\text{b}}^{r}} where expansion is (a + b)n{{\left( a\text{ + b} \right)}^{n}} and lastly use the identity nCr = nCn−r{}^{n}{{C}_{r}}\text{ = }{}^{n}{{C}_{n-r}}

Complete step-by-step solution:
In the question we are asked to find the coefficient of x6{{x}^{6}} in [(1+x)6+(1+x)7+....(1+x)15]\left[ {{\left( 1+x \right)}^{6}}+{{\left( 1+x \right)}^{7}}+....{{\left( 1+x \right)}^{15}} \right]
At first let’s analyze this sum which is (1 + x)6 + (1 + x)7 + (1 + x)8 + . . . . . . + (1 + x)15{{\left( 1\text{ + }x \right)}^{6}}\text{ + }{{\left( 1\text{ + }x \right)}^{7}}\text{ + }{{\left( 1\text{ + }x \right)}^{8}}\text{ + }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. + }{{\left( 1\text{ + }x \right)}^{15}}
So, the given series is in geometric progression.
Here first term (a) of geometric progression is (1 + x)6{{\left( 1\text{ + }x \right)}^{6}} and its common ratio can be found out by finding ratio the successive term by preceding term, so, it will be (1+x)7(1+x)6⇒(1+x)\dfrac{{{\left( 1+x \right)}^{7}}}{{{\left( 1+x \right)}^{6}}}\Rightarrow \left( 1+x \right)
Now to find the sum we use formula,
Sum=a (rn - 1)(r - 1)\text{Sum}=\dfrac{a\,({{r}^{n}}\text{ - 1)}}{(r\text{ - 1)}}………. (1)
To solve the above equation, we need the value of “a, r and n”. We know the value of “a and r” but don’t know the value of “n” which we are going to calculate below:
There is a formula for nth{{n}^{th}} term of G.P. which is equal to:
Tn=arn−1{{T}_{n}}=a{{r}^{n-1}}
Substituting Tn=(1+x)15,a=(1+x)6,r=(1+x){{T}_{n}}={{\left( 1+x \right)}^{15}},a={{\left( 1+x \right)}^{6}},r=\left( 1+x \right) in the above equation we get,
(1+x)15=(1+x)6(1+x)n−1 ⇒(1+x)15=(1+x)6+n−1 ⇒(1+x)15=(1+x)5+n \begin{aligned} & {{\left( 1+x \right)}^{15}}={{\left( 1+x \right)}^{6}}{{\left( 1+x \right)}^{n-1}} \\\ & \Rightarrow {{\left( 1+x \right)}^{15}}={{\left( 1+x \right)}^{6+n-1}} \\\ & \Rightarrow {{\left( 1+x \right)}^{15}}={{\left( 1+x \right)}^{5+n}} \\\ \end{aligned}
Now, the base is the same on both the sides so we can equate the powers.
15=5+n ⇒10=n \begin{aligned} & 15=5+n \\\ & \Rightarrow 10=n \\\ \end{aligned}
Now, we got the value of n too so we can find the summation of the series.
So, here “a” is first term, “r” is common ratio and “n” is number of terms of the series so substituting the values that we solved above in (1) we get,
Sum=a (rn - 1)(r - 1)\text{Sum}=\dfrac{a\,({{r}^{n}}\text{ - 1)}}{(r\text{ - 1)}}
⇒(1 + x)6((1 + x)10 - 1)((1 + x) - 1)\Rightarrow \dfrac{{{\left( 1\text{ + }x \right)}^{6}}\left( {{\left( 1\text{ + }x \right)}^{10}}\text{ - 1} \right)}{\left( \left( 1\text{ + }x \right)\text{ - 1} \right)}
which can be simplified as

& \dfrac{{{\left( 1\text{ + }x \right)}^{6}}\left( {{\left( 1\text{ + }x \right)}^{10}}\text{ - 1} \right)}{x} \\\ & \Rightarrow \dfrac{{{\left( 1\text{ + }x \right)}^{16}}-{{\left( 1\text{ + }x \right)}^{6}}}{x} \\\ \end{aligned}$$ It can be written as $$\dfrac{{{\left( 1\text{ + }x \right)}^{16}}}{x}\text{ - }\dfrac{{{\left( 1\text{ + }x \right)}^{6}}}{x}$$ As we have to find the coefficient of ${{x}^{6}}$ but as there is variable x in the denominator so we have to find the coefficient of ${{x}^{7}}$ which is not possible for the ${{2}^{nd}}$ term as the highest power of ${{\left( 1\text{ + }x \right)}^{6}}$ is ${{x}^{6}}$ and when divided by x it becomes ${{x}^{5}}$ which is not asked. Hence we have to find the expansion of ${{x}^{7}}$ in the binomial ${{\left( 1\text{ + }x \right)}^{16}}$ Let the general term of binomial expansion be $${{T}_{r+1}}\text{ = }{}^{n}{{C}_{r}}\text{ }{{\text{1}}^{n-r}}\text{ }{{x}^{r}}$$ using formula $${{T}_{r+1}}\text{ = }{}^{n}{{C}_{r}}\text{ }{{\text{a}}^{n-r}}\text{ }{{\text{b}}^{r}}$$ if the expression is $${{\left( a\text{ + b} \right)}^{n}}$$ So the term $${{T}_{r+1}}\text{ = }{}^{n}{{C}_{r}}\text{ }{{\left( \text{1} \right)}^{n-r}}\text{ }{{x}^{r}}$$ $$\Rightarrow {{T}_{r+1}}\text{ = }{}^{n}{{C}_{r}}{{x}^{r}}$$ In the question we are already told that the exponent of $x$ is $7$ so the value of r is $$7$$. Here the value of n is 16. So the $${{T}_{8}}\text{ = }{}^{16}{{C}_{7}}\text{ }{{x}^{7}}$$ Hence the coefficient of ${{x}^{7}}$ in $${{\left( 1\text{ + }x \right)}^{16}}$$ is ${}^{16}{{C}_{7}}$ Now as we know that there is an identity, $${}^{n}{{C}_{r}}\text{ = }{}^{n}{{C}_{n-r}}$$ where n is 16 and r is 7. So, $${}^{16}{{C}_{7}}\text{ = }{}^{16}{{C}_{16-7}}\Rightarrow {}^{16}{{C}_{7}}\text{ = }{}^{16}{{C}_{9}}$$ **Hence the correct option is ‘A’.** **Note:** One can also do by just expanding each and every term by using binomial expansion and then adding all the terms of coefficient of ${{x}^{6}}$ to get the answer as this process is long and very tedious it is preferred not to do.