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Question: If \[{\left( {2 + \dfrac{x}{3}} \right)^{55}}\]is expanded in the ascending powers of x in two conse...

If (2+x3)55{\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. 7th{7^{th}}and 8th{8^{th}}
B. 8th{8^{th}}and 9th{9^{th}}
C. 28th{28^{th}}and 29th{29^{th}}
D. 27th{27^{th}}and 28th{28^{th}}

Explanation

Solution

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{(a + b)^n} through the formula (a+b)n=∑r=0nnCr(a)n−r(b)r{(a + b)^n} = \sum\limits_{r = 0}^n {^n{C_r}{{(a)}^{n - r}}{{(b)}^r}}

Complete step-by-step answer:
We can expand (2+x3)55{\left( {2 + \dfrac{x}{3}} \right)^{55}}using binomial expansion where a=2,b=x3,n=55a = 2,b = \dfrac{x}{3},n = 55.
Now we take two consecutive terms Pr+1,Pr+2{P_{r + 1}},{P_{r + 2}}
We can write (r+1)th{(r + 1)^{th}}term as Pr+1=55Cr(2)55−r(x3)r{P_{r + 1}}{ = ^{55}}{C_r}{(2)^{55 - r}}{(\dfrac{x}{3})^r}
Similarly we can write (r+2)th{(r + 2)^{th}}term as Pr+2=55Cr+1(2)55−r−1(x3)r+1{P_{r + 2}}{ = ^{55}}{C_{r + 1}}{(2)^{55 - r - 1}}{(\dfrac{x}{3})^{r + 1}}
i.e. Pr+2=55Cr+1(2)54−r(x3)r+1{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 nCr=n!(n−r)!(r)!^n{C_r} = \dfrac{{n!}}{{(n - r)!(r)!}}
Coefficient of Pr+1{P_{r + 1}}term is 55Cr(2)55−r(13)r^{55}{C_r}{(2)^{55 - r}}{(\dfrac{1}{3})^r}
⇒55Cr(2)55−r(13)r=55!(55−r)!(r)!(2)55−r(13)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 Pr+2{P_{r + 2}}term is 55Cr+1(2)54−r(13)r+1^{55}{C_{r + 1}}{(2)^{54 - r}}{(\dfrac{1}{3})^{r + 1}}
⇒55Cr+1(2)54−r(13)r+1=55!(55−(r+1))!(r+1)!(2)54−r(13)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}}
⇒55Cr+1(2)54−r(13)r+1=55!(54−r)!(r+1)!(2)54−r(13)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 Pr+1,Pr+2{P_{r + 1}},{P_{r + 2}}.
⇒55Cr(2)55−r(13)r=55Cr+1(2)54−r(13)r+1{ \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)
⇒55!(55−r)!(r)!(2)55−r(13)r=55!(54−r)!(r+1)!(2)54−r(13)r+1\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)!(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)!(55 - r)! = (55 - r)(55 - r - 1)! = (55 - r)(54 - r)!
(r+1)!=(r+1)(r)!(r + 1)! = (r + 1)(r)!
And using the rule of exponents am+n=aman{a^{m + n}} = {a^m}{a^n} we write
(2)55−r=(2)1+54−r=(2)(2)54−r{(2)^{55 - r}} = {(2)^{1 + 54 - r}} = (2){(2)^{54 - r}}
(13)r+1=(13)r(13){(\dfrac{1}{3})^{r + 1}} = {(\dfrac{1}{3})^r}(\dfrac{1}{3})
Substituting the values in equation (3)

⇒55!(55−r)(55−r−1)!(r)!(2)1+54−r(13)r=55!(54−r)!(r+1)(r)!(2)54−r(13)r(13) ⇒55!(55−r)(54−r)!(r)!(2)(2)54−r(13)r=55!(54−r)!(r+1)(r)!(2)54−r(13)r(13)  \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
⇒1(55−r)(2)=1(r+1)(13)\Rightarrow \dfrac{1}{{(55 - r)}}(2) = \dfrac{1}{{(r + 1)}}(\dfrac{1}{3})
Cross multiply the terms on both sides

⇒3×2×(r+1)=55−r ⇒6(r+1)=55−r ⇒6r+6=55−r  \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

⇒6r+r=55−6 ⇒7r=49  \Rightarrow 6r + r = 55 - 6 \\\ \Rightarrow 7r = 49 \\\

Divide both sides by 7
⇒7r7=497\Rightarrow \dfrac{{7r}}{7} = \dfrac{{49}}{7}
Cancel out terms from numerator and denominator
⇒r=7\Rightarrow r = 7
Substituting the value of r in Pr+1,Pr+2{P_{r + 1}},{P_{r + 2}} we get the two consecutive terms as P8,P9{P_8},{P_9}
Therefore, 8th{8^{th}} and 9th{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)!(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.