Advertisements
Advertisements
प्रश्न
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 expaand]
Advertisements
उत्तर
a = a − b + b
So, an = [a − b +b]n
= [(a − b) + b]n
= nC0 (a − b)n + nC1 (a − b)n−1b1 + nC2 (a − b)n−2b2 + ....... + nCn−1 (a − b)bn−1 + nCn (bn)
⇒ an − bn = (a − b)n + nC1 (a − b)n−1b + nC2 (a − b)n−2b2 + ....... + nCn−1 (a − b)bn−1
= (a − b) [(a − b)n−1 + nC1 (a − b)n−2b + nC2 (a − b)n−3b2 + ...... + nCn−1 bn−1]
= (a – b)[an integer]
⇒ an – bn is divisible by (a – b)
APPEARS IN
संबंधित प्रश्न
Expand the following by using binomial theorem.
(2a – 3b)4
Expand the following by using binomial theorem.
`(x + 1/y)^7`
Expand the following by using binomial theorem.
`(x + 1/x^2)^6`
Find the 5th term in the expansion of (x – 2y)13.
Find the middle terms in the expansion of
`(x + 1/x)^11`
Find the middle terms in the expansion of
`(3x + x^2/2)^8`
Prove that the term independent of x in the expansion of `(x + 1/x)^(2n)` is `(1*3*5...(2n - 1)2^n)/(n!)`.
Show that the middle term in the expansion of is (1 + x)2n is `(1*3*5...(2n - 1)2^nx^n)/(n!)`
Find the Co-efficient of x11 in the expansion of `(x + 2/x^2)^17`
The middle term in the expansion of `(x + 1/x)^10` is
The constant term in the expansion of `(x + 2/x)^6` is
Sum of the binomial coefficients is
Expand `(2x^2 -3sqrt(1 - x^2))^4 + (2x^2 + 3sqrt(1 - x^2))^4`
Compute 994
If n is a positive integer and r is a non-negative integer, prove that the coefficients of xr and xn−r in the expansion of (1 + x)n are equal
In the binomial expansion of (1 + x)n, the coefficients of the 5th, 6th and 7th terms are in AP. Find all values of n
Choose the correct alternative:
The remainder when 3815 is divided by 13 is
