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