Advertisements
Advertisements
Question
Evaluate the following using binomial theorem:
(999)5
Solution
(x + a)n = nC0 xn a0 + nC1 xn-1 a1 + nC2 xn-2 a2 + ……… + nCr xn-r ar + …… + nCn an
(999)5 = (1000 – 1)5 = 5C0 (1000)5 – 5C1 (1000)4 (1)1 + 5C2 (1000)3 (1)2 – 5C3 (1000)2 (1)3 + 5C4 (1000)5 (1)4 – 5C5 (1)5
= 1(1000)5 – 5(1000)4 – 10(1000)3 – 10(1000)2 + 5(1000) – 1
= 1000000000000000 – 5000000000000 + 10000000000 – 10000000 + 5000 – 1
= 995009990004999
APPEARS IN
RELATED QUESTIONS
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`
The middle term in the expansion of `(x + 1/x)^10` is
The constant term in the expansion of `(x + 2/x)^6` is
Find the coefficient of x15 in `(x^2 + 1/x^3)^10`
Find the coefficient of x4 in the expansion `(1 + x^3)^50 (x^2 + 1/x)^5`
Find the last two digits of the number 3600
If n is an odd positive integer, prove that the coefficients of the middle terms in the expansion of (x + y)n are equal
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]
Choose the correct alternative:
The value of 2 + 4 + 6 + … + 2n is
