Question

The greatest terms of the expansion $ {\left( {2x + 5y} \right)^{13}} $ when x = 10, y = 2 is
A) $ ^{13}{C_5}{.20^8}{.10^4} $
B) $ ^{13}{C_6}{.20^7}{.10^5} $
C) $ ^{13}{C_4}{.20^9}{.10^4} $
D) None of these

Answer 1

Hint: As the power of the given expression is very large, we can find its greatest term using the general formula for the binomial expansion. While expanding, we can substitute the given values of x and y so as to find the value of r and get the required greatest term.
Formula to be used:
For \[{\left( {x + y} \right)^n}\], the general term is given as:
\[{T_{r + 1}}{ = ^n}{C_r}.{x^{n - r}}.{y^r}\]
The value of $ ^n{C_r} $ can be calculated using the factorial formula as:
 $ ^n{C_r} = \dfrac{{n!}}{{(n - r)!r!}} $

Complete step by step solution:
We can find the expansion of the given term using the binomial expansion. But, we need to find the greatest term of expression, for that, we will use the formula for the general term of binomial expansion given as:
\[
  {\left( {x + y} \right)^n}: \\
   \Rightarrow {T_{r + 1}}{ = ^n}{C_r}.{x^{n - r}}.{y^r}....(1) \\
 \]
If $ {T_r} $ be the largest term of this expansion, it will lie between the other terms as:
 $ {T_{r - 1}} < {T_r} > {T_{r + 1}}....(2) $
The base of the general term is (r+1), writing the bases of these terms in this context, we get:
 $ {T_{r - 2 + 1}} < {T_{r - 1 + 1}} > {T_{r + 1}} $
Expressions of these terms using (1) can be given as:
 $
  {T_{r - 1}} \Rightarrow {T_{r - 2 + 1}}{ = ^n}{C_{r - 2}}.{x^{n - \left( {r - 2} \right)}}.{y^{r - 2}} \\
  {T_r} \Rightarrow {T_{r - 1 + 1}}{ = ^n}{C_{r - 1}}.{x^{n - \left( {r - 1} \right)}}.{y^{r - 1}} \\
  {T_{r + 1}} \Rightarrow {T_r}{ = ^n}{C_r}.{x^{n - r}}.{y^r} \;
  $
Now, the given expression is $ {\left( {2x + 5y} \right)^{13}} $ , comparing it with $ {\left( {x + y} \right)^n} $ :
x = 2x
y = 5y
n = 13
Substituting these values in the obtained expressions, we get:
 $
  {T_{r - 1}}{ = ^{13}}{C_{r - 2}}.{\left( {2x} \right)^{13 - \left( {r - 2} \right)}}.{\left( {5y} \right)^{r - 2}} \\
   \Rightarrow {T_{r - 1}}{ = ^{13}}{C_{r - 2}}.{\left( {2x} \right)^{15 - r}}.{\left( {5y} \right)^{r - 2}} \\
  {T_r}{ = ^{13}}{C_{r - 1}}.{\left( {2x} \right)^{13 - \left( {r - 1} \right)}}.{\left( {5y} \right)^{r - 1}} \\
   \Rightarrow {T_r}{ = ^{13}}{C_{r - 1}}.{\left( {2x} \right)^{14 - r}}.{\left( {5y} \right)^{r - 1}} \\
  {T_{r + 1}}{ = ^{13}}{C_r}.{\left( {2x} \right)^{13 - r}}.{\left( {5y} \right)^r} \\
  $
From (2):
\[
  {T_{r - 1}} < {T_r} \\
   \Rightarrow \dfrac{{{T_{r - 1}}}}{{{T_r}}} < 1 \\
   \Rightarrow \dfrac{{{T_r}}}{{{T_{r - 1}}}} > 1 \\
 \]
Substituting the values:
 $ \dfrac{{^{13}{C_{r - 1}}{{\left( {2x} \right)}^{14 - r}}{{\left( {5y} \right)}^{r - 1}}}}{{^{13}{C_{r - 2}}{{\left( {2x} \right)}^{15 - r}}{{\left( {5y} \right)}^{r - 2}}}} > 1 $
Using $ ^n{C_r} = \dfrac{{n!}}{{(n - r)!r!}} $ , the inequality becomes:
\[
\dfrac{{\dfrac{{13!}}{{\left[ {13 - \left( {r - 1} \right)} \right]!\left( {r - 1} \right)!}}}}{{\dfrac{{13!}}{{\left[ {13 - \left( {r - 2} \right)} \right]!\left( {r - 2} \right)!}}}} \times {\left( {2x} \right)^{14 - r - 15 + r}} \times {\left( {5y} \right)^{r - 1 - r + 2}} > 1 \\
\Rightarrow \dfrac{{\dfrac{{13!}}{{\left( {14 - r} \right)!\left( {r - 1} \right)!}}}}{{\dfrac{{13!}}{{\left( {15 - r} \right)!\left( {r - 2} \right)!}}}} \times \dfrac{1}{{2x}} \times 5y > 1 \\
   \Rightarrow \dfrac{{15 - r}}{{r - 1}} \times \dfrac{1}{{2x}} \times 5y > 1 \;
 \]y
It is given that x = 10 and y=2, so:
 $
   \Rightarrow \dfrac{{15 - r}}{{r - 1}} \times \dfrac{1}{{20}} \times 10 > 1 \\
   \Rightarrow 15 - r > 2r - 2 \\
   \Rightarrow 3r < 17 \\
   \Rightarrow r < \dfrac{{17}}{3} \\
   \Rightarrow r < 5.6....(i) \;
  $
From (2):
\[
  {T_r} > {T_{r + 1}} \\
   \Rightarrow \dfrac{{{T_r}}}{{{T_{r + 1}}}} > 1 \\
   \Rightarrow \dfrac{{{T_{r + 1}}}}{{{T_r}}} < 1 \;
 \]
Substituting the values:
 $ \dfrac{{^{13}{C_r}{{\left( {2x} \right)}^{13 - r}}{{\left( {5y} \right)}^r}}}{{^{13}{C_{r - 1}}{{\left( {2x} \right)}^{14 - r}}{{\left( {5y} \right)}^{r - 1}}}} < 1 $
Using $ ^n{C_r} = \dfrac{{n!}}{{(n - r)!r!}} $ , the inequality becomes:
\[
  \dfrac{{\dfrac{{13!}}{{\left( {13 - r} \right)!r!}}}}{{\dfrac{{13!}}{{\left[ {13 - \left( {r - 1} \right)} \right]!\left( {r - 1} \right)!}}}} \times {\left( {2x} \right)^{13 - r - 14 + r}} \times {\left( {5y} \right)^{r - r + 1}} < 1 \\
\Rightarrow \dfrac{{\dfrac{{13!}}{{\left( {13 - r} \right)!r!}}}}{{\dfrac{{13!}}{{\left( {14 - r} \right)!\left( {r - 1} \right)!}}}} \times \dfrac{1}{{2x}} \times 5y < 1 \\
  \Rightarrow \dfrac{{14 - r}}{{2r}} \times \dfrac{1}{{2x}} \times 5y < 1 \;
 \]
It is given that x = 10 and y=2, so:
 $
   \Rightarrow \dfrac{{14 - r}}{{2r}} \times \dfrac{1}{{20}} \times 10 < 1 \\
   \Rightarrow 14 - r > 2r \\
   \Rightarrow 3r > 14 \\
   \Rightarrow r > \dfrac{{14}}{3} \\
   \Rightarrow r > 4.6....(ii) \;
  $
The value of r lies between 5.6 and 4.6, so the value of 5 as a whole number can be 5
So, the value the largest term will be given as:
 $
  {T_5} = {T_{4 + 1}} \\
  {T_{4 + 1}}{ = ^{13}}{C_4}{\left( {2x} \right)^9}{\left( {5y} \right)^4} \\
  {T_5}{ = ^{13}}{C_4}{\left( {20} \right)^9}{\left( {10} \right)^4}\left[ {\because x = 10,y = 2} \right] \;
  $
Therefore, The greatest terms of the expansion $ {\left( {2x + 5y} \right)^{13}} $ when x = 10, y = 2 is $ ^{13}{C_4}{.20^9}{.10^4} $
So, the correct answer is “Option C”.

Note: When we have greater than or smaller than sign between RHS and LHS values, the equation is known as inequality, when we reciprocal, we change the sign as well. In the question,
When \[\dfrac{{{T_r}}}{{{T_{r + 1}}}} > 1\] was given, but we required its reciprocal, so the sign also changes which give:
\[\dfrac{{{T_{r + 1}}}}{{{T_r}}} < 1\]
For the calculations, remember that for the same base, the powers are added and subtracted when the bases are multiplied and divided respectively.
We use the ’!’ sign to represent the factorial which means the product from 1 to the particular number.