Asked by Sinachi
Write the four terms of the binomial expansion (2-1/x)^5 in ascending powers of X. Hence find (1.9)^5
Answers
Answered by
Tammy
Must be too hard for the AI to handle ....
(2-1/x)^5
= 2^5 + 5(2^4)(-1/x) + 10(2^3)(-1/x)^2 + 10(2^2)(-1/x)^3 + ..
= 32 - 80/x + 80/x^2 - 40/x^3 + ...
in (1.9)^5, we would have (2 - .1)^5 , -1/x = -.1, 1/x = 1/10, x = 10
so
(1.9^5) = 32 - 80/10 + 80/100 - 40/1000
= 32 - 8 + .8 - .04 +
= 24.76
actual value of 1.9^5 using calculator = 24.76099...
(2-1/x)^5
= 2^5 + 5(2^4)(-1/x) + 10(2^3)(-1/x)^2 + 10(2^2)(-1/x)^3 + ..
= 32 - 80/x + 80/x^2 - 40/x^3 + ...
in (1.9)^5, we would have (2 - .1)^5 , -1/x = -.1, 1/x = 1/10, x = 10
so
(1.9^5) = 32 - 80/10 + 80/100 - 40/1000
= 32 - 8 + .8 - .04 +
= 24.76
actual value of 1.9^5 using calculator = 24.76099...
Answered by
oobleck
looks like your expansion is in decreasing powers of x -- that is, increasing powers of 1/x
One expansion is
1 - ∑ (1-x)^n
One expansion is
1 - ∑ (1-x)^n
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