Advertisements
Advertisements
Question
Using binomial theorem, indicate which of the following two number is larger: `(1.01)^(1000000)`, 10
Advertisements
Solution
`(1.01)^(1000000) = (1 + 0.01)^(1000000)`
= `""^(1000000)"C"_0(1)^(1000000) + ""^(1000000)"C"_1(1)^(999999)(0.01)^1 + ""^(1000000)"C"_2 (1)^(999998)(0.01)^2 + ""^(1000000)"C"_3(1)^(999997)(0.01)^3 + ..........`
= `1(1) + 1000000xx 1/10^2 + (1000000 xx 999999)/2 xx 1/10000 + .........`
= 1 + 10000 + 50 × 999999 + ........ which is > 10000
So `(1.01)^(1000000) > 10000`
(i.e.) `(1.01)^(1000000)` is larger
APPEARS IN
RELATED QUESTIONS
Evaluate the following using binomial theorem:
(999)5
Expand the following by using binomial theorem.
(2a – 3b)4
Expand the following by using binomial theorem.
`(x + 1/y)^7`
Find the middle terms in the expansion of
`(x + 1/x)^11`
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 constant term in the expansion of `(x + 2/x)^6` is
The last term in the expansion of (3 + √2 )8 is:
Sum of binomial coefficient in a particular expansion is 256, then number of terms in the expansion is:
Sum of the binomial coefficients is
Expand `(2x^2 -3sqrt(1 - x^2))^4 + (2x^2 + 3sqrt(1 - x^2))^4`
Compute 1024
Compute 994
Compute 97
Find the coefficient of x2 and the coefficient of x6 in `(x^2 -1/x^3)^6`
Find the last two digits of the number 3600
If n is a positive integer, using Binomial theorem, show that, 9n+1 − 8n − 9 is always divisible by 64
Prove that `"C"_0^2 + "C"_1^2 + "C"_2^2 + ... + "C"_"n"^2 = (2"n"!)/("n"!)^2`
