Question

31. The numerically greatest term in the expansion of (3 + 2x)49 when x =1/5
1) 4th term
2) 5 th term
3) 6th term
4) 7th term
O 1
2
ОО
3
4​

Answer 1

The (r + 1)th term in the expansion of (3 + 2x)⁴⁹ is,

$\small\longrightarrow T_{r+1}=\,^{49}\!C_r\ 3^{49-r}\ (2x)^r$

Taking x = 1/5,

$\small\longrightarrow T_{r+1}=\,^{49}\!C_r\ 3^{49-r}\ \left(\dfrac{2}{5}\right)^r$

Let (r + 1)th term be numerically the greatest term, 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 r'th term is numerically less than or equal to (r + 1)th term (there can be two adjacent terms being numerically the greatest of all terms but numerically equal to each other), i.e.,

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

$\small\longrightarrow\left|\,^{49}\!C_{r-1}\ 3^{49-(r-1)}\ \left(\dfrac{2}{5}\right)^{r-1}\right|\leq\left|\,^{49}\!C_r\ 3^{49-r}\ \left(\dfrac{2}{5}\right)^r\right|$

Modulus can be removed since each term is positive.

$\small\longrightarrow\,^{49}\!C_{r-1}\ 3^{50-r}\ \left(\dfrac{2}{5}\right)^{r-1}\leq\,^{49}\!C_r\ 3^{49-r}\ \left(\dfrac{2}{5}\right)^r$

$\small\longrightarrow\dfrac{49!}{(r-1)!(49-r+1)!}\cdot3^{50-r}\ \left(\dfrac{2}{5}\right)^{r-1}\leq\dfrac{49!}{r!(49-r)!}\cdot3^{49-r}\ \left(\dfrac{2}{5}\right)^r$

$\small\longrightarrow\dfrac{49!}{(r-1)!(49-r)!(49-r+1)}\cdot3\cdot3^{49-r}\ \left(\dfrac{2}{5}\right)^{r-1}\leq\dfrac{49!}{r(r-1)!(49-r)!}\cdot3^{49-r}\ \left(\dfrac{2}{5}\right)^{r-1}\cdot\dfrac{2}{5}$

Cancelling like terms,

$\small\longrightarrow\dfrac{1}{49-r+1}\cdot3\leq\dfrac{1}{r}\cdot\dfrac{2}{5}$

$\small\longrightarrow\dfrac{3}{50-r}\leq\dfrac{2}{5r}$

$\small\longrightarrow15r\leq100-2r$

$\small\longrightarrow17r\leq100$

$\small\longrightarrow r\leq\dfrac{100}{17}$

Since r is a whole number,

$\small\longrightarrow r\leq5$

The numerically greatest term is given by r = 5 so (r + 1)th = 6th term is numerically the greatest. Other values of r or r ≤ 5 means the first 6 terms in the expansion are in ascending order.

Hence (3) is the answer.

Answer 2

Answer:

the answer is 6th term.

option c)

Hope it helps

#Be Brainly.