Advertisements
Advertisements
Question
Find the 4th term from the end in the expansion of `(x^3/2 - 2/x^2)^9`
Advertisements
Solution
Since rth term from the end in the expansion of (a + b)n is (n – r + 2)th term from the beginning.
Therefore 4th term from the end is 9 – 4 + 2
i.e., 7th term from the beginning
Which is given by T7 = `""^9"C"_6 (x^3/2) ((-2)/x^2)^6`
= `""^9"C"_3 x^9/8 * 64/x^12`
= `(9 xx 8 xx 7)/(3 xx 2 xx 1) xx 64/x^3`
= `672/x^3`
APPEARS IN
RELATED QUESTIONS
Expand the expression (1– 2x)5
Using binomial theorem, evaluate f the following:
(101)4
Using binomial theorem, evaluate the following:
(99)5
Find (x + 1)6 + (x – 1)6. Hence or otherwise evaluate `(sqrt2 + 1)^6 + (sqrt2 -1)^6`
Show that 9n+1 – 8n – 9 is divisible by 64, whenever n is a positive integer.
Find a, b and n in the expansion of (a + b)n if the first three terms of the expansion are 729, 7290 and 30375, respectively.
Find a if the coefficients of x2 and x3 in the expansion of (3 + ax)9 are equal.
Evaluate `(sqrt3 + sqrt2)^6 - (sqrt3 - sqrt2)^6`
Find the value of `(a^2 + sqrt(a^2 - 1))^4 + (a^2 - sqrt(a^2 -1))^4`
Expand using Binomial Theorem `(1+ x/2 - 2/x)^4, x != 0`
Find the value of (1.01)10 + (1 − 0.01)10 correct to 7 places of decimal.
Show that \[2^{4n + 4} - 15n - 16\] , where n ∈ \[\mathbb{N}\] is divisible by 225.
Expand the following (1 – x + x2)4
Determine whether the expansion of `(x^2 - 2/x)^18` will contain a term containing x10?
Show that `2^(4n + 4) - 15n - 16`, where n ∈ N is divisible by 225.
If a1, a2, a3 and a4 are the coefficient of any four consecutive terms in the expansion of (1 + x)n, prove that `(a_1)/(a_1 + a_2) + (a_3)/(a_3 + a_4) = (2a_2)/(a_2 + a_3)`
If (1 – x + x2)n = a0 + a1 x + a2 x2 + ... + a2n x2n , then a0 + a2 + a4 + ... + a2n equals ______.
The number of terms in the expansion of (a + b + c)n, where n ∈ N is ______.
Find the coefficient of x in the expansion of (1 – 3x + 7x2)(1 – x)16.
Find the coefficient of x15 in the expansion of (x – x2)10.
If the coefficient of second, third and fourth terms in the expansion of (1 + x)2n are in A.P. Show that 2n2 – 9n + 7 = 0.
Find the coefficient of x4 in the expansion of (1 + x + x2 + x3)11.
The total number of terms in the expansion of (x + a)100 + (x – a)100 after simplification is ______.
Given the integers r > 1, n > 2, and coefficients of (3r)th and (r + 2)nd terms in the binomial expansion of (1 + x)2n are equal, then ______.
The two successive terms in the expansion of (1 + x)24 whose coefficients are in the ratio 1:4 are ______.
The coefficient of a–6b4 in the expansion of `(1/a - (2b)/3)^10` is ______.
Let the coefficients of x–1 and x–3 in the expansion of `(2x^(1/5) - 1/x^(1/5))^15`, x > 0, be m and n respectively. If r is a positive integer such that mn2 = 15Cr, 2r, then the value of r is equal to ______.
If the coefficients of (2r + 4)th, (r – 2)th terms in the expansion of (1 + x)18 are equal, then r is ______.
Let `(5 + 2sqrt(6))^n` = p + f where n∈N and p∈N and 0 < f < 1 then the value of f2 – f + pf – p is ______.
