Advertisements
Advertisements
Question
If 3rd, 4th 5th and 6th terms in the expansion of (x + a)n be respectively a, b, c and d, prove that `(b^2 - ac)/(c^2 - bd) = (5a)/(3c)`.
Advertisements
Solution
Given expansion is (x + a)n.
`"T"_3 = a = ^nC_2 x^(n - 2) a^2 = (n(n - 1))/2 x^(n - 2) a^2`
`"T"_4 = b = ^nC_3 x^(n - 3) a^3 = (n(n - 1)(n - 2))/6 x^(n - 3) a^3`
`"T"_5 = c = ^nC_4 x^(n - 4) a^4 = (n(n - 1)(n - 2)(n - 3))/24 x^(n - 4) a^4`
`"T"_6 = d = ^nC_5 x^(n - 5) a^5 = (n(n - 1)(n - 2)(n - 3)(n - 4))/120 x^(n - 5) a^5`
Now,
`("T"_4)/("T"_3) = b/a = [(n(n - 1)(n - 2))/6 × x^(n - 3) × a^3]/[(n(n - 1))/2 × x^(n - 2) × a^2] = (n - 2)/3 . a/x ...(1)`
`("T"_5)/("T"_4) = c/b = [(n(n - 1)(n - 2)(n - 3))/24 × x^(n - 4) × a^4]/[(n(n - 1)(n - 2))/6 × x^(n - 3) × a^3] = (n - 3)/4 . a/x ...(2)`
`("T"_6)/("T"_5) = d/c = [(n(n - 1)(n - 2)(n-3)(n-4))/120 × x^(n - 5) × a^5]/[(n(n - 1)(n - 2)(n - 3))/24 × x^(n - 4) × a^4] = (n - 4)/5 . a/x ...(3)`
Again, dividing (1) by (2) and (2) by (3), we get
`[("T"_4)/("T"_3)]/[("T"_5)/("T"_4)] = [b/a]/[c/b] = [(n - 2)/3 . a/x]/[(n - 3)/4 . a/x] = [4(n - 2)]/[3(n - 3)]`
⇒ `(b^2)/(ac) = [4(n - 2)]/[3(n - 3)] ...(4)`
and
`[("T"_5)/("T"_4)]/[("T"_6)/("T"_5)] = [c/b]/[d/c] = [(n - 3)/4 . a/x]/[(n - 4)/5 . a/x] = [5(n- 3)]/[4(n - 4)].`
⇒ `[c^2]/[bd] = [5(n- 3)]/[4(n - 4)] ...(5)`
Now subtact 1 from both sides of equation (4) and (5) as:
⇒ `(b^2)/(ac) - 1 = [4(n - 2)]/[3(n - 3)] - 1`
⇒ `(b^2 - ac)/(ac) = (n + 1)/(3(n - 3))` ...(6)
and
⇒ `[c^2]/[bd] - 1 = [5(n- 3)]/[4(n - 4)] - 1`
⇒ `[c^2 - bd]/[bd] = [(n + 1)]/[4(n - 4)]` ...(7)
Again, on dividing (6) by (7), we get
`[(b^2 - ac)/(ac)]/[[c^2 - bd]/[bd]] = [(n + 1)/(3(n - 3))]/[[(n + 1)]/[4(n - 4)]]`
`(b^2 - ac)/(c^2 - bd) × (bd)/(ac) = [4(n - 4)]/[3(n - 3)]` ...(8)
On multiplying (5) by (8),
`(b^2 − ac)/(c^2 − bd) × (bd)/(ac) × c^2/(bd) = [4(n − 4)]/[3(n − 3)] × [5(n − 3)]/[4(n − 4)]`
⇒ `(b^2 − ac)/(c^2 − bd).c/a = 5/3`
⇒ `(b^2 - ac)/(c^2 - bd) = (5a)/(3c)`.
Hence proved.
APPEARS IN
RELATED QUESTIONS
Write the general term in the expansion of (x2 – y)6
Find the 4th term in the expansion of (x – 2y)12 .
In the expansion of (1 + a)m + n, prove that coefficients of am and an are equal.
The coefficients of the (r – 1)th, rth and (r + 1)th terms in the expansion of (x + 1)n are in the ratio 1:3:5. Find n and r.
Find the middle term in the expansion of:
(i) \[\left( \frac{2}{3}x - \frac{3}{2x} \right)^{20}\]
Find the middle term in the expansion of:
(iv) \[\left( \frac{x}{a} - \frac{a}{x} \right)^{10}\]
Find the middle terms in the expansion of:
(iii) \[\left( 3x - \frac{2}{x^2} \right)^{15}\]
Find the middle terms(s) in the expansion of:
(i) \[\left( x - \frac{1}{x} \right)^{10}\]
Find the term independent of x in the expansion of the expression:
(i) \[\left( \frac{3}{2} x^2 - \frac{1}{3x} \right)^9\]
Find the term independent of x in the expansion of the expression:
(ix) \[\left( \sqrt[3]{x} + \frac{1}{2 \sqrt[3]{x}} \right)^{18} , x > 0\]
If the coefficients of (2r + 1)th term and (r + 2)th term in the expansion of (1 + x)43 are equal, find r.
Prove that the coefficient of (r + 1)th term in the expansion of (1 + x)n + 1 is equal to the sum of the coefficients of rth and (r + 1)th terms in the expansion of (1 + x)n.
The coefficients of 5th, 6th and 7th terms in the expansion of (1 + x)n are in A.P., find n.
If in the expansion of (1 + x)n, the coefficients of pth and qth terms are equal, prove that p + q = n + 2, where \[p \neq q\]
Find the coefficient of a4 in the product (1 + 2a)4 (2 − a)5 using binomial theorem.
In the expansion of (1 + x)n the binomial coefficients of three consecutive terms are respectively 220, 495 and 792, find the value of n.
If in the expansion of (1 + x)n, the coefficients of three consecutive terms are 56, 70 and 56, then find n and the position of the terms of these coefficients.
If the 6th, 7th and 8th terms in the expansion of (x + a)n are respectively 112, 7 and 1/4, find x, a, n.
Find a, b and n in the expansion of (a + b)n, if the first three terms in the expansion are 729, 7290 and 30375 respectively.
If p is a real number and if the middle term in the expansion of \[\left( \frac{p}{2} + 2 \right)^8\] is 1120, find p.
Write the middle term in the expansion of `((2x^2)/3 + 3/(2x)^2)^10`.
Write the coefficient of the middle term in the expansion of \[\left( 1 + x \right)^{2n}\] .
Write the total number of terms in the expansion of \[\left( x + a \right)^{100} + \left( x - a \right)^{100}\] .
Find the middle term in the expansion of `(2ax - b/x^2)^12`.
Find the term independent of x, x ≠ 0, in the expansion of `((3x^2)/2 - 1/(3x))^15`
If the term free from x in the expansion of `(sqrt(x) - k/x^2)^10` is 405, find the value of k.
Find the middle term (terms) in the expansion of `(3x - x^3/6)^9`
Find the coefficient of `1/x^17` in the expansion of `(x^4 - 1/x^3)^15`
Find the value of r, if the coefficients of (2r + 4)th and (r – 2)th terms in the expansion of (1 + x)18 are equal.
If the middle term of `(1/x + x sin x)^10` is equal to `7 7/8`, then value of x is ______.
The position of the term independent of x in the expansion of `(sqrt(x/3) + 3/(2x^2))^10` is ______.
The coefficient of x256 in the expansion of (1 – x)101(x2 + x + 1)100 is ______.
Let for the 9th term in the binomial expansion of (3 + 6x)n, in the increasing powers of 6x, to be the greatest for x = `3/2`, the least value of n is n0. If k is the ratio of the coefficient of x6 to the coefficient of x3, then k + n0 is equal to ______.
Let the coefficients of the middle terms in the expansion of `(1/sqrt(6) + βx)^4, (1 - 3βx)^2` and `(1 - β/2x)^6, β > 0`, common difference of this A.P., then `50 - (2d)/β^2` is equal to ______.
