Question

The numerically greatest terms in the expansion of ${(2x+5y)}^{34}$ when $x=3$ & $y=2$ is

• A. $T_{21}$
• B. $T_{22}$
• C. $T_{23}$
• D. $T_{24}$

Answer 1

The correct option is A $T_{21}$

$(2x+5y{)}^{34}$ can be written as $(2x{)}^{34}$.$(1+\frac{5y}{2x}{)}^{34}$

$\to$ or $(2x{)}^{34}$.$(1+\alpha {)}^{34}$, where $\alpha$ = $\frac{5y}{3x}$

Value of $\alpha$ at x=3, y=2 is equal to $\frac{5.2}{2.3}=\frac{5}{3}$

So ${\alpha }_{x=3,y=2}=\frac{5}{3}$

Now the numerically greatest term in the expansion of $(2x+5y{)}^{34}$ is the numerically greatest term in the expansion of $(1+\alpha {)}^{34}$

We know that for any expansion $(1+\alpha {)}^n$, If the numerically greatest term is $T_r$, where r is an integer.

Then the value of r is $\le \frac{(n+1)|\alpha |}{|\alpha +1|}$

So to know the numerically greatest term in the given expansion at x=3, y=2 we need to find the integer value of
r $\le \frac{(n+1)|\alpha |}{|\alpha +1|}$ at x=3, y=2, Here the value of n = 34

So $r\le \frac{(34+1)|\frac{5}{3}|}{|\frac{5}{3}+1|}$

$\to$ $\frac{35.5}{8}=\frac{175}{8}=21.8$

As r is an integer so the closest value of integer less than 21.8 is 21.

Hence the numerically greatest term will be 21th term or $T_{21}$

So the correct option is A.