Question

If x= 5/2, numerically the greatest term in the expansion of (3+2x)^15 is​

Answer 1

The (r + 1)th term in the expansion of (3 + 2x)¹⁵ is,

$\small\longrightarrow T_{r+1}=\,^{15}\!C_r\ 3^{15-r}\ (2x)^r$

Taking x = 5/2,

$\small\longrightarrow T_{r+1}=\,^{15}\!C_r\ 3^{15-r}\ \left(2\cdot\dfrac{5}{2}\right)^r$

$\small\longrightarrow T_{r+1}=\,^{15}\!C_r\ 3^{15-r}\ 5^r$

Let (r + 1)th term be numerically the greatest term, means let the magnitude of (r + 1)th term be the greatest, so not only the magnitude of every terms before and after the (r + 1)th term are lesser, but also the terms before this term are in ascending order and the terms after this term are in descending order.

Then we can say r'th term is numerically less than or equal to (r + 1)th term (there can be two adjacent terms being numerically the greatest but having same magnitude), i.e.,

$\small\longrightarrow|T_r|\leq|T_{r+1}|$

$\small\longrightarrow\left|\,^{15}\!C_{r-1}\ 3^{15-(r-1)}\ 5^{r-1}\right|\leq\left|\,^{15}\!C_r\ 3^{15-r}\ 5^r\right|$

Modulus can be removed since each term is positive.

$\small\longrightarrow\,^{15}\!C_{r-1}\ 3^{16-r}\ 5^{r-1}\leq\,^{15}\!C_r\ 3^{15-r}\ 5^r$

$\small\longrightarrow\dfrac{15!}{(r-1)!(15-r+1)!}\cdot3^{16-r}\ 5^{r-1}\leq\dfrac{15!}{r!(15-r)!}\cdot3^{15-r}\ 5^r$

$\small\longrightarrow\dfrac{15!}{(r-1)!(15-r)!(15-r+1)}\cdot3\cdot3^{15-r}\ 5^{r-1}\leq\dfrac{15!}{r(r-1)!(15-r)!}\cdot3^{15-r}\ 5^{r-1}\cdot5$

Cancelling like terms,

$\small\longrightarrow\dfrac{1}{15-r+1}\cdot3\leq\dfrac{1}{r}\cdot5$

$\small\longrightarrow\dfrac{3}{16-r}\leq\dfrac{5}{r}$

$\small\longrightarrow3r\leq80-5r$

$\small\longrightarrow8r\leq80$

$\small\longrightarrow r\leq10$

The numerically greatest term is given by r = 10 so (r + 1)th = 11th term is numerically the greatest. Other values of r or r ≤ 10 means the first 11 terms in the expansion are in ascending order.

So the numerically greatest term is,

$\small\longrightarrow T_{11}=\,^{15}\!C_{10}\ 3^{15-10}\ 5^{10}$

$\small\longrightarrow T_{11}=\dfrac{15!}{10!\cdot5!}\cdot3^5\cdot5^{10}$

$\small\text{ \longrightarrow\underline{\underline{T_{11}=71,26,25,97,65,625}} }$