1.

The numerically greatest term in the expansion (2x-3y)¹² when x=1 and y= 5/2 is​

Answer»

Given:

(2X-3y)¹² , x=1 and y= 5/2  

To find:

The numerically GREATEST term in the expansion (2x-3y)¹² when x=1 and y= 5/2 is​

Solution:

From given, we have,

(2x-3y)¹², x=1 and y= 5/2  

we USE the formula,

P = [ (n + 1) |x| ] / [|x| + 1]  

to find the numerically greatest term.

so we have,

(2x-3y)¹² = (2x)¹²(1 - 3y/2x)¹²

Thus we GET,

|x| = (3/2) (y/x)

substituting the values of x and y, we get,

|x| = (3/2) (5/2) = 15/4

Now consider,

P = [ (n + 1)|x| ] / [|x| + 1]  

P = [ (12 + 1) (15/4)] / [15/4 + 1]  

P = [ (13) (15/4)] / [19/4]  

P = 195/19 = 10.26

Thus 10.26 ≈ 11, 11TH term is the numerically greatest term in the expansion (2x-3y)¹² when  x=1 and y= 5/2.


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