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Answer:
To determine the greatest coefficient in the binomial expansion, (1+x)n, when n is a positive integer. Coefficient of
(Tr+1/Tr) = Cr/Cr-1 = (n-r+1)/r = ((n+1)/r) - 1.
Now the (r+1)th binomial coefficient will be greater than the rth binomial coefficient when, Tr+1 > Tr
=> ((n+1)/r)-1 > 1 => (n+1)/2 >r. ....... (1)
But r must be an integer, and therefore when n is even, the greatest binomial coefficient is given by the greatest value of r, consistent with (1) i.e., r = n/2 and hence the greatest binomial coefficient is nCn/2.
Similarly in n be odd, the greatest binomial coefficient is given when,
r = (n-1)/2 or (n+1)/2 and the coefficient itself will be nC(n+1)/2 or nC(n-1)/2, both being are equal
Note: The greatest binomial coefficient is the binomial coefficient of the middle term.
Illustration:
Show that the greatest the coefficient in the expansion of (x + 1/x)2n is (1.3.5...(2n-1).2n)/n! .
Solution:
Since middle term has the greatest coefficient.
So, greatest coefficient = coefficient of middle term
= 2nCn = (1.2.3...2n)/n!n! = (1.3.5...(2n-1).2n)/n!.
Numerically greatest term
To determine the numerically greatest term in the expansion of (a + x)n, where n is a positive integer.
Consider
numerically-greatest-term
Thus
numerically-greatest-term-eqn
Note : {((n+1)/r) - 1} must be positive since n > r. Thus Tr+1 will be the greatest term if, r has the greatest value as per the equation (1).
Illustration:
Find the greatest term in the expansion of (3-2x)9 when x = 1.
Solution:
Tr+1/Tr = ((9-r+1)/r) . (2x/3) >1
i.e. 20 > 5r
If r = 4, then Tr+1 = Tr and these are the greatest terms. Thus 4th and 5th terms are numerically equal and greater than any other term and their value is equal 39 × 9C3 × (2/3)3 = 489888.
Illustration:
Find the greatest term in the expansion of (2 + 3x)9 if x = 3/2.
Solution:
Here Tr+1 /Tr = ((n-r+1)/r)(3x/2) = ((10-r)/r)(3x/2), (where x = 3/2)
= ((10-r)/r)(3x/2)(3/2) = ((10-r)/r).9/4 = (90-9r)/4r
Therefore Tr+1 > Tr, if 90 - 9r > 4r.
=> 90 > 13r => r < 90/13 and r being an integer, r = 6.
Hence Tr+1 = T7 = T6+1 = 9C6 (2)3 (3x)6 = 313.7/2.
Answer:
To determine the greatest coefficient in the binomial expansion, (1+x)n, when n is a positive integer. Coefficient of
(Tr+1/Tr) = Cr/Cr-1 = (n-r+1)/r = ((n+1)/r) - 1.
Now the (r+1)th binomial coefficient will be greater than the rth binomial coefficient when, Tr+1 > Tr
=> ((n+1)/r)-1 > 1 => (n+1)/2 >r. ....... (1)
But r must be an integer, and therefore when n is even, the greatest binomial coefficient is given by the greatest value of r, consistent with (1) i.e., r = n/2 and hence the greatest binomial coefficient is nCn/2.
Similarly in n be odd, the greatest binomial coefficient is given when,