Advertisements
Advertisements
Question
Evaluate the following using binomial theorem:
(999)5
Advertisements
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
Evaluate the following using binomial theorem:
(101)4
Find the middle terms in the expansion of
`(x + 1/x)^11`
Find the middle terms in the expansion of
`(2x^2 - 3/x^3)^10`
Find the term independent of x in the expansion of
`(2x^2 + 1/x)^12`
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!)`
The last term in the expansion of (3 + √2 )8 is:
Find the coefficient of x2 and the coefficient of x6 in `(x^2 -1/x^3)^6`
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
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]
