Question

The coefficient of ${x^9}$ in the polynomial given by $\sum\nolimits_{r = 1}^{11} {\left( {x + r} \right)\left( {x + r + 1} \right)\left( {x + r + 2} \right)...\left( {x + r + 9} \right)}$ is
(a)5511
(b)5151
(c)1515
(d)1155

Answer 1

Hint:- Here, we will be using the concept of coefficients in the expansion of multiplication of n linear terms and then we will be using the formula of summation of n terms of an arithmetic progression

Complete step-by-step answer:
It has given in the question that we need to find out the coefficient. The coefficient of ${x^9}$ in the polynomial $\sum\nolimits_{r = 1}^{11} {\left( {x + r} \right)\left( {x + r + 1} \right)\left( {x + r + 2} \right)...\left( {x + r + 9} \right)}$
We can now rewrite the given expression
$\Rightarrow$ coefficient of ${x^9}$ in the polynomial $\sum\nolimits_{r = 1}^{11} {\left( {x + r} \right)\left( {x + r + 1} \right)\left( {x + r + 2} \right)...\left( {x + r + 9} \right)}$ by taking the summation sign outside as
$\Rightarrow$ $\sum\nolimits_{r = 1}^{11} {coefficient\,\,of\,\,{x^9}\,\,in\,\,\left( {x + r} \right)\left( {x + r + 1} \right)\left( {x + r + 2} \right)...\left( {x + r + 9} \right)}$……….(i)
Now let us study the given polynomial $\left( {x + r} \right)\left( {x + r + 1} \right)\left( {x + r + 2} \right)...\left( {x + r + 9} \right)$
We can observe that it can be written as $\left( {x + \left( r \right)} \right)\left( {x + \left( {r + 1} \right)} \right)\left( {x + \left( {r + 2} \right)} \right)...\left( {x + \left( {r + 9} \right)} \right)$, which can further be written as $\left( {x + a} \right)\left( {x + b} \right)\left( {x + c} \right)...\left( {x + d} \right)$ where $a = r,b = r + 1,c = r + 2,...,d = r + 9$
Since there are 10 terms in the polynomial $\left( {x + a} \right)\left( {x + b} \right)\left( {x + c} \right)...\left( {x + d} \right)$, the highest power of x is 10, which is even.
We can now say that the second term in the expansion of the given polynomial with have the variable x raised to the power of 9.
Since the highest power is even, then the coefficient of the next highest power of x , which in this case will be 9) will be equal to the positive of the sum of the roots.
In this case, the sum of the roots is $a + b + c + ... + d$ which is equal to
$\left( r \right) + \left( {r + 1} \right) + \left( {r + 2} \right) + ... + \left( {r + 9} \right)$
Thus we get $\left( r \right) + \left( {r + 1} \right) + \left( {r + 2} \right) + ... + \left( {r + 9} \right)$ as the coefficient of ${x^9}$ in $\left( {x + r} \right)\left( {x + r + 1} \right)\left( {x + r + 2} \right)...\left( {x + r + 9} \right)$
Equation (i) can be written as
$\Rightarrow$ $\sum\nolimits_{r = 1}^{11} {\left[ {r + \left( {r + 1} \right) + \left( {r + 2} \right) + \left( {r + 3} \right) + ... + \left( {r + 9} \right)} \right]}$
We can further simplify it by adding the terms within the summation. to get
$\Rightarrow \sum\nolimits_{r = 1}^{11} {\left[ {10r + \left( {1 + 2 + 3 + ... + 9} \right)} \right]}$
Separating out the terms we get

\Rightarrow 10\sum\nolimits_{r = 1}^{11} r + \sum\nolimits_{r = 1}^{11} {\left[ {\left( {1 + 2 + 3 + ... + 9} \right)} \right]} \\\ \Rightarrow 10\sum\nolimits_{r = 1}^{11} r + 45\sum\nolimits_{r = 1}^{11} 1 \end{array}$$ Expanding the summation we get We can now use the summation formula of n terms in an AP, i.e ${S_n} = \dfrac{n}{2}\left\\{ {2a + \left( {n - 1} \right)d} \right\\}$ , to get $$\begin{array}{l} \Rightarrow 10 \times \dfrac{{11 \times 12}}{2} + 45 \times 11\\\ \Rightarrow 660 + 495\\\ \Rightarrow 1155 \end{array}$$ Note:- We must try to find out the coefficient of ${x^9}$in the given polynomial by using the concept of coefficients of different powers of the variable in expansion of the given polynomial. Once we have done that, then only terms of r will remain within the summation sign and the question can easily be solved after that, using summation of arithmetic progression.