Advertisements
Advertisements
प्रश्न
The coefficient of a–6b4 in the expansion of `(1/a - (2b)/3)^10` is ______.
Advertisements
उत्तर
The coefficient of a–6b4 in the expansion of `(1/a - (2b)/3)^10` is `1120/27`.
Explanation:
The given expansion is `(1/a - (2b)/3)^10`
From a–6b4
We can take r = 4
∴ T5 = T4+1
= `""^10"C"_4 (1/a)^(10 - 4) (- (2b)/3)^4`
= `""^10"C"_4 (1/a)^6 ((-2)/3)^4 * b^4`
= `(10*9*8*7)/(4*3*2*1) xx 16/81 * a^-6b^4`
= `210 xx 16/81 a^-6b^4`
= `1120/27 a^-6b^4`
APPEARS IN
संबंधित प्रश्न
Expand the expression: (1– 2x)5
Expand the expression (1– 2x)5
Expand the expression: `(2/x - x/2)^5`
Expand the expression: `(x + 1/x)^6`
Using Binomial Theorem, evaluate the following:
(96)3
Using Binomial Theorem, evaluate of the following:
(102)5
Using binomial theorem, evaluate the following:
(99)5
Using Binomial Theorem, indicate which number is larger (1.1)10000 or 1000.
Find (a + b)4 – (a – b)4. Hence, evaluate `(sqrt3 + sqrt2)^4 - (sqrt3 - sqrt2)^4`
Find (x + 1)6 + (x – 1)6. Hence or otherwise evaluate `(sqrt2 + 1)^6 + (sqrt2 -1)^6`
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.
Find the coefficient of x5 in the product (1 + 2x)6 (1 – x)7 using binomial theorem.
If a and b are distinct integers, prove that a – b is a factor of an – bn, whenever n is a positive integer.
[Hint: write an = (a – b + b)n and expand]
Find an approximation of (0.99)5 using the first three terms of its expansion.
Expand using Binomial Theorem `(1+ x/2 - 2/x)^4, x != 0`
Expand the following (1 – x + x2)4
If n is a positive integer, find the coefficient of x–1 in the expansion of `(1 + x)^2 (1 + 1/x)^n`
Which of the following is larger? 9950 + 10050 or 10150
The total number of terms in the expansion of (x + a)51 – (x – a)51 after simplification is ______.
If the coefficients of x7 and x8 in `2 + x^n/3` are equal, then n is ______.
If (1 – x + x2)n = a0 + a1 x + a2 x2 + ... + a2n x2n , then a0 + a2 + a4 + ... + a2n equals ______.
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.
In the expansion of (x + a)n if the sum of odd terms is denoted by O and the sum of even term by E. Then prove that 4OE = (x + a)2n – (x – a)2n
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 ______.
Number of terms in the expansion of (a + b)n where n ∈ N is one less than the power n.
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 ______.
The sum of the last eight coefficients in the expansion of (1 + x)16 is equal to ______.
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 ______.
