Advertisements
Advertisements
प्रश्न
Find the sixth term of the expansion `(y^(1/2) + x^(1/3))^"n"`, if the binomial coefficient of the third term from the end is 45.
Advertisements
उत्तर
The given expression is `(y^(1/2) + x^(1/3))^"n"`
Since the binomial coefficient of third term from the end = Binomial coefficient of third term from the beginning = nC2
∴ nC2 = 45
⇒ `("n"("n" - 1))/2` = 45
⇒ n2 – n = 90
⇒ n2 – n – 90 = 0
⇒ n2 – 10n + 9n – 90 = 0
⇒ n(n – 10) + 9(n – 10) = 0
⇒ (n – 10)(n + 9) = 0
⇒ n = 10, n = –9
⇒ n = 10, n ≠ – 9
So, the given expression becomes `(y^(1/2) + x^(1/3))^10`
Sixth term is this expression T6 = T5+1
= `""^10"C"_5 (y^(1/2))^(10 - 5) (x^(1/3))^5`
= `""^10"C"_5 y^(5/2) * x^(5/3)`
= `252 y^(5/2) x^(5/3)`
Hence, the required term = `252 y^(5/2) * x^(5/3)`
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: (2x – 3)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
Find (a + b)4 – (a – b)4. Hence, evaluate `(sqrt3 + sqrt2)^4 - (sqrt3 - sqrt2)^4`
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.
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]
Evaluate `(sqrt3 + sqrt2)^6 - (sqrt3 - sqrt2)^6`
Find an approximation of (0.99)5 using the first three terms of its expansion.
Find the expansion of (3x2 – 2ax + 3a2)3 using binomial theorem.
Show that \[2^{4n + 4} - 15n - 16\] , where n ∈ \[\mathbb{N}\] is divisible by 225.
Find the 4th term from the end in the expansion of `(x^3/2 - 2/x^2)^9`
Evaluate: `(x^2 - sqrt(1 - x^2))^4 + (x^2 + sqrt(1 - x^2))^4`
Show that `2^(4n + 4) - 15n - 16`, where n ∈ N is divisible by 225.
Which of the following is larger? 9950 + 10050 or 10150
If (1 – x + x2)n = a0 + a1 x + a2 x2 + ... + a2n x2n , then a0 + a2 + a4 + ... + a2n equals ______.
The coefficient of xp and xq (p and q are positive integers) in the expansion of (1 + x)p + q are ______.
The number of terms in the expansion of (a + b + c)n, where n ∈ N is ______.
If z = `sqrt(3)/2 + i^5/2 + sqrt(3)/2 - i^5/2`, then ______.
Find the coefficient of x in the expansion of (1 – 3x + 7x2)(1 – x)16.
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.
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 O2 – E2 = (x2 – a2)n
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
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 ______.
The sum of the last eight coefficients in the expansion of (1 + x)16 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 ______.
The positive integer just greater than (1 + 0.0001)10000 is ______.
