Question

If \[{\left( {2 + \dfrac{x}{3}} \right)^{55}}\]is expanded in the ascending powers of x in two consecutive terms of the expansion are equal then these terms are:
A. \[{7^{th}}\]and \[{8^{th}}\]
B. \[{8^{th}}\]and \[{9^{th}}\]
C. \[{28^{th}}\]and \[{29^{th}}\]
D. \[{27^{th}}\]and \[{28^{th}}\]

Answer 1

Hint: We use the method of binomial expansion to expand the term in the bracket in ascending order of x. Then using the method to write a term of binomial expansion we write two consecutive terms and equate their coefficients.
* A binomial expansion helps us to expand expressions of the form \[{(a + b)^n}\] through the formula \[{(a + b)^n} = \sum\limits_{r = 0}^n {^n{C_r}{{(a)}^{n - r}}{{(b)}^r}} \]

Complete step-by-step answer:
 We can expand \[{\left( {2 + \dfrac{x}{3}} \right)^{55}}\]using binomial expansion where \[a = 2,b = \dfrac{x}{3},n = 55\].
Now we take two consecutive terms \[{P_{r + 1}},{P_{r + 2}}\]
We can write \[{(r + 1)^{th}}\]term as \[{P_{r + 1}}{ = ^{55}}{C_r}{(2)^{55 - r}}{(\dfrac{x}{3})^r}\]
Similarly we can write \[{(r + 2)^{th}}\]term as \[{P_{r + 2}}{ = ^{55}}{C_{r + 1}}{(2)^{55 - r - 1}}{(\dfrac{x}{3})^{r + 1}}\]
i.e. \[{P_{r + 2}}{ = ^{55}}{C_{r + 1}}{(2)^{54 - r}}{(\dfrac{x}{3})^{r + 1}}\]
We solve the coefficient by using the formula for combination \[^n{C_r} = \dfrac{{n!}}{{(n - r)!(r)!}}\]
Coefficient of \[{P_{r + 1}}\]term is \[^{55}{C_r}{(2)^{55 - r}}{(\dfrac{1}{3})^r}\]
\[{ \Rightarrow ^{55}}{C_r}{(2)^{55 - r}}{(\dfrac{1}{3})^r} = \dfrac{{55!}}{{(55 - r)!(r)!}}{(2)^{55 - r}}{(\dfrac{1}{3})^r}\] … (1)
Coefficient of \[{P_{r + 2}}\]term is \[^{55}{C_{r + 1}}{(2)^{54 - r}}{(\dfrac{1}{3})^{r + 1}}\]
\[{ \Rightarrow ^{55}}{C_{r + 1}}{(2)^{54 - r}}{(\dfrac{1}{3})^{r + 1}} = \dfrac{{55!}}{{(55 - (r + 1))!(r + 1)!}}{(2)^{54 - r}}{(\dfrac{1}{3})^{r + 1}}\]
\[{ \Rightarrow ^{55}}{C_{r + 1}}{(2)^{54 - r}}{(\dfrac{1}{3})^{r + 1}} = \dfrac{{55!}}{{(54 - r)!(r + 1)!}}{(2)^{54 - r}}{(\dfrac{1}{3})^{r + 1}}\] … (2)
Now we know the coefficients of the consecutive terms are equal. So we equate the coefficients of terms \[{P_{r + 1}},{P_{r + 2}}\].
\[{ \Rightarrow ^{55}}{C_r}{(2)^{55 - r}}{(\dfrac{1}{3})^r}{ = ^{55}}{C_{r + 1}}{(2)^{54 - r}}{(\dfrac{1}{3})^{r + 1}}\]
Substitute the values from equation (1) and (2)
\[ \Rightarrow \dfrac{{55!}}{{(55 - r)!(r)!}}{(2)^{55 - r}}{(\dfrac{1}{3})^r} = \dfrac{{55!}}{{(54 - r)!(r + 1)!}}{(2)^{54 - r}}{(\dfrac{1}{3})^{r + 1}}\] … (3)
Now using \[(n + 1)! = (n + 1)(n)!\] we expand the terms of factorial on both sides and write \[(55 - r)! = (55 - r)(55 - r - 1)! = (55 - r)(54 - r)!\]
\[(r + 1)! = (r + 1)(r)!\]
And using the rule of exponents \[{a^{m + n}} = {a^m}{a^n}\] we write
\[{(2)^{55 - r}} = {(2)^{1 + 54 - r}} = (2){(2)^{54 - r}}\]
\[{(\dfrac{1}{3})^{r + 1}} = {(\dfrac{1}{3})^r}(\dfrac{1}{3})\]
Substituting the values in equation (3)
\[
   \Rightarrow \dfrac{{55!}}{{(55 - r)(55 - r - 1)!(r)!}}{(2)^{1 + 54 - r}}{(\dfrac{1}{3})^r} = \dfrac{{55!}}{{(54 - r)!(r + 1)(r)!}}{(2)^{54 - r}}{(\dfrac{1}{3})^r}(\dfrac{1}{3}) \\
   \Rightarrow \dfrac{{55!}}{{(55 - r)(54 - r)!(r)!}}(2){(2)^{54 - r}}{(\dfrac{1}{3})^r} = \dfrac{{55!}}{{(54 - r)!(r + 1)(r)!}}{(2)^{54 - r}}{(\dfrac{1}{3})^r}(\dfrac{1}{3}) \\
 \]
Cancel out same terms from both sides of the equations
\[ \Rightarrow \dfrac{1}{{(55 - r)}}(2) = \dfrac{1}{{(r + 1)}}(\dfrac{1}{3})\]
Cross multiply the terms on both sides
\[
   \Rightarrow 3 \times 2 \times (r + 1) = 55 - r \\
   \Rightarrow 6(r + 1) = 55 - r \\
   \Rightarrow 6r + 6 = 55 - r \\
 \]
Shift the terms with variable on one side and constants on other side of the equation
\[
   \Rightarrow 6r + r = 55 - 6 \\
   \Rightarrow 7r = 49 \\
 \]
Divide both sides by 7
\[ \Rightarrow \dfrac{{7r}}{7} = \dfrac{{49}}{7}\]
Cancel out terms from numerator and denominator
\[ \Rightarrow r = 7\]
Substituting the value of r in \[{P_{r + 1}},{P_{r + 2}}\] we get the two consecutive terms as \[{P_8},{P_9}\]
Therefore, \[{8^{th}}\] and \[{9^{th}}\] terms are having equal coefficients
So, option B is correct.

Note: Students are likely to make mistake while writing the factorial into simpler form as they tend to make mistake of writing \[(55 - r)! = (55 - r)(55 - (r - 1))! = (55 - r)(56 - r)!\] which is wrong, we have to subtract 1 from whole term inside the bracket.