Question

The third, fourth and fifth terms in the expansion of ${\left( {x + a} \right)^n}$ in descending power of x are 84, 280 and 560 respectively, find x, a and n.
$(a){\text{ x = - 1, a = - 2, n = 2}} \\\ (b){\text{ x = 1, a = - 2, n = - 2}} \\\ (c){\text{ x = - 1, a = 2, n = - 2}} \\\ (d){\text{ x = 1, a = 2, n = 2}} \\\$

Answer 1

Hint – In this question use the direct formula for any rth term a binomial expansion of ${\left( {x + a} \right)^n}$ that is ${T_{r + 1}} = {}^n{C_r}{x^{n - r}}{a^r}$, write the 4th, 5th and 3rd term in this expansion using this formula to get the values of a, x and n.

Complete step-by-step answer:
According to binomial expansion of ${\left( {x + a} \right)^n}$ the general term
${T_{r + 1}} = {}^n{C_r}{x^{n - r}}{a^r}$, where $0 \leqslant r \leqslant n$ so when we increases the value of r the power of x is decreases.
Therefore 3rd, 4th and 5th term in the expansion of ${\left( {x + a} \right)^n}$ is
${T_3} = {}^n{C_2}{x^{n - 2}}{a^2} = 84$............................... (1)
Similarly,
${T_4} = {}^n{C_3}{x^{n - 3}}{a^3} = 280$............................... (2)
${T_5} = {}^n{C_4}{x^{n - 4}}{a^4} = 560$............................... (3)
Now divide equation (2) from (1) we have,
$\Rightarrow \dfrac{{{}^n{C_3}{x^{n - 3}}{a^3}}}{{{}^n{C_2}{x^{n - 2}}{a^2}}} = \dfrac{{280}}{{84}}$
Now simplify the above equation we have,
$\Rightarrow \dfrac{{{}^n{C_3}{a^{3 - 2}}}}{{{}^n{C_2}{x^{n - 2 - n + 3}}}} = \dfrac{{280}}{{84}}$
$\Rightarrow \dfrac{{{}^n{C_3}}}{{{}^n{C_2}}}\left( {\dfrac{a}{x}} \right) = \dfrac{{10}}{3}$......................... (4)
Now divide equation (3) by (2) we have,
$\Rightarrow \dfrac{{{}^n{C_4}{x^{n - 4}}{a^4}}}{{{}^n{C_3}{x^{n - 3}}{a^3}}} = \dfrac{{560}}{{280}} = 2$
$\Rightarrow \dfrac{{{}^n{C_4}}}{{{}^n{C_3}}}\left( {\dfrac{a}{x}} \right) = 2$............................... (5)
Now as we know that ${}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$ so use this property we have,
From equation (4) we have,
$\Rightarrow \dfrac{{\dfrac{{n!}}{{3!\left( {n - 3} \right)!}}}}{{\dfrac{{n!}}{{2!\left( {n - 2} \right)!}}}}\left( {\dfrac{a}{x}} \right) = \dfrac{{10}}{3}$
$\Rightarrow \dfrac{{n!}}{{3!\left( {n - 3} \right)!}} \times \dfrac{{2!\left( {n - 2} \right)!}}{{n!}}\left( {\dfrac{a}{x}} \right) = \dfrac{{10}}{3}$
$\Rightarrow \dfrac{{\left( {n - 2} \right)}}{3}\left( {\dfrac{a}{x}} \right) = \dfrac{{10}}{3}$.................. (6)
From equation (5) we have,
$\Rightarrow \dfrac{{{}^n{C_4}}}{{{}^n{C_3}}}\left( {\dfrac{a}{x}} \right) = 2$
$\Rightarrow \dfrac{{\dfrac{{n!}}{{4!\left( {n - 4} \right)!}}}}{{\dfrac{{n!}}{{3!\left( {n - 3} \right)!}}}}\left( {\dfrac{a}{x}} \right) = 2$
$\Rightarrow \dfrac{{n!}}{{4!\left( {n - 4} \right)!}} \times \dfrac{{3!\left( {n - 3} \right)!}}{{n!}}\left( {\dfrac{a}{x}} \right) = 2$
$\Rightarrow \dfrac{{\left( {n - 3} \right)}}{4}\left( {\dfrac{a}{x}} \right) = 2$....................... (7)
Now divide equation (6) by equation (7) we have,
$\Rightarrow \dfrac{{\dfrac{{\left( {n - 2} \right)}}{3}\left( {\dfrac{a}{x}} \right)}}{{\dfrac{{\left( {n - 3} \right)}}{4}\left( {\dfrac{a}{x}} \right)}} = \dfrac{{\dfrac{{10}}{3}}}{2}$
Now simplify the above equation we have,
$\Rightarrow \dfrac{{4\left( {n - 2} \right)}}{{3\left( {n - 3} \right)}} = \dfrac{5}{3}$
$\Rightarrow 4n - 8 = 5n - 15$
$\Rightarrow 5n - 4n = 15 - 8$
$\Rightarrow n = 7$
Now from equation (6) we have,
$\Rightarrow \dfrac{{\left( {7 - 2} \right)}}{3}\left( {\dfrac{a}{x}} \right) = \dfrac{{10}}{3}$
$\Rightarrow \dfrac{a}{x} = 2$
$\Rightarrow a = 2x$............. (8)
Now from equation (1) we have,
$\Rightarrow {}^7{C_2}{x^{7 - 2}}{a^2} = 84$
$\Rightarrow \dfrac{{7!}}{{2!.5!}}{x^5}{a^2} = 84$
$\Rightarrow \dfrac{{7 \times 6 \times 5!}}{{2 \times 1 \times 5!}}{x^5}{a^2} = 84$
$\Rightarrow 21{x^5}{a^2} = 84$
$\Rightarrow {x^5}{a^2} = 4$
Now from equation (8) we have,
$\Rightarrow {x^5}{\left( {2x} \right)^2} = 4$
$\Rightarrow 4{x^7} = 4$
$\Rightarrow {x^7} = 1$
$\Rightarrow x = 1$
Now from equation (8) we have
$\Rightarrow a = 2x = 2\left( 1 \right) = 2$
So the required values a = 2, x= 1 and n= 7.
So this is the required answer.
Hence option (D) is correct.

Note – The binomial theorem is an algebraic method of expanding a binomial expression, essentially it demonstrates what happens when we multiply a binomial by itself as many times as we need. It is advised to remember some of the basic direct formulas for binomial expansion like the general term of an expansion as it helps save a lot of time.