Expand (1+2x)^20 - (1-2x)^20) in ascending powers of x up to the term in x^5 and use this result to evaluate (1.02)^20 - (0.98)^20 to 3 significant figures.
(1+2x)^20 - (1-2x)^20)
=(1 + 20(2x) + 190(4x^2 + 1140(8x^3) + 4845(16x^4) + 15504(32x^5 + ...) - (1 - 20(2x) + 190(4x^2) - 1140(8x^3) + 4845(16x^4) - 15504(32x^5) + ..)
= 80x + 18240x^3 + 992256x^5
comparing (1.02)^20 - (0.98)^20 with (1+2x)^20 - (1-2x)^20
we can see that x = .01
so if x = .01
80x + 18240x^3 + 992256x^5
= 80(.01) + 18240(.00001) + 992256(.000000001)
= .8 +01824 + .0000992256
= .818339225
= .818 using 3 significant figures
real answer: .818339425
As I said before, a rather futile exercise, merely showing the mathematical property but not at all practical.
If x had not been such a nice number like .01, but rather something like x = .31 we would still have a horrible calculation to do
In the expansion of (1-2x)^11 the coefficient of x^3 is k times the coefficient of x^2. Evaluate k.
To expand the expression (1+2x)^20 - (1-2x)^20, we will use the binomial theorem. The binomial theorem states that for any positive integer n:
(a + b)^n = C(n, 0) * a^n * b^0 + C(n, 1) * a^(n-1) * b^1 + C(n, 2) * a^(n-2) * b^2 + ... + C(n, n-2) * a^2 * b^(n-2) + C(n, n-1) * a^1 * b^(n-1) + C(n, n) * a^0 * b^n
Where C(n, k) represents the binomial coefficient which is equal to n! / (k!(n-k)!), where ! denotes factorial.
Now, let's apply the binomial theorem to expand (1+2x)^20 and (1-2x)^20:
(1+2x)^20 = C(20, 0) * 1^20 * (2x)^0 + C(20, 1) * 1^19 * (2x)^1 + C(20, 2) * 1^18 * (2x)^2 + ... + C(20, 18) * 1^2 * (2x)^18 + C(20, 19) * 1^1 * (2x)^19 + C(20, 20) * 1^0 * (2x)^20
(1-2x)^20 = C(20, 0) * 1^20 * (-2x)^0 + C(20, 1) * 1^19 * (-2x)^1 + C(20, 2) * 1^18 * (-2x)^2 + ... + C(20, 18) * 1^2 * (-2x)^18 + C(20, 19) * 1^1 * (-2x)^19 + C(20, 20) * 1^0 * (-2x)^20
Now, let's simplify each term in the above two expansions:
(1+2x)^20 = 1 + 40x + 380x^2 + 1520x^3 + 3420x^4 + 4845x^5 + ...
(1-2x)^20 = 1 - 40x + 380x^2 - 1520x^3 + 3420x^4 - 4845x^5 + ...
To find the expansion of (1+2x)^20 - (1-2x)^20, we subtract the corresponding terms in the two expansions:
(1+2x)^20 - (1-2x)^20 = (1 + 40x + 380x^2 + 1520x^3 + 3420x^4 + 4845x^5 + ...) - (1 - 40x + 380x^2 - 1520x^3 + 3420x^4 - 4845x^5 + ...)
Simplifying further, we get:
(1+2x)^20 - (1-2x)^20 = 1 + 40x + 2 * (380x^2 + 1520x^3 + 3420x^4 + 4845x^5) + ...
Now, we only need to consider the terms up to x^5. Hence, the expansion in ascending powers of x up to the term in x^5 is:
(1+2x)^20 - (1-2x)^20 = 1 + 40x + 760x^2 + 3040x^3 + 6840x^4 + 9690x^5 + ...
Next, we'll use this expansion to evaluate (1.02)^20 - (0.98)^20 to 3 significant figures.
Substitute x = 0.01 into the expansion:
(1+2x)^20 - (1-2x)^20 = 1 + 40(0.01) + 760(0.01)^2 + 3040(0.01)^3 + 6840(0.01)^4 + 9690(0.01)^5 + ...
Now, perform the calculations:
(1+2x)^20 - (1-2x)^20 = 1 + 0.4 + 0.076 + 0.00304 + 0.0000684 + 0.0000969 + ...
Evaluate the expression to 3 significant figures:
(1+2x)^20 - (1-2x)^20 = 1.480
Therefore, (1.02)^20 - (0.98)^20 evaluates to 1.480 to 3 significant figures.