Find the first 3 terms in the expansion of (2x - 3/x)^5 in descending powers of x. Then find the coefficient of x in the expansion of (1+ 2/x^2)(2x - 3/x)^5.
Please help me solve this problem, 10 points??
Please help me solve this problem, 10 points??
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Hello,
You should use the binomial distribution that states that, for any reals a and b, and any positive integer n:
(a + b)ⁿ = ∑(p=0→n) {n! / [p! × (n - p)!)]} × aⁿ⁻ᵖ × bᵖ
Thus is your case, with a=2x, b=3/x and n=5
(2x - 3/x)⁵ =
5!/(0! × 5!) × (2x)⁵(-3/x)º
+ 5!/(1! × 4!) × (2x)⁴(-3/x)¹
+ 5!/(2! × 3!) × (2x)³(-3/x)²
+ 5!/(3! × 2!) × (2x)²(-3/x)³
+ 5!/(4! × 1!) × (2x)¹(-3/x)⁴
+ 5!/(5! × 0!) × (2x)º(-3/x)⁵
= 1 × 32x⁵
+ 5 × (-48x³)
+ 10 × 72x
+ 10 × (-108/x)
+ 5 × 162/x³
+ 1 × (-243/x⁵)
= 32x⁵ - 240x³ + 720x - 1080/x + 810/x³ - 243/x⁵
Thus the three first terms in descending powers of (2x-3/x)⁵ are:
32x⁵, -240x³ and 720x.
The only two ways to obtain x is:
1 × 720x = 720x and
-240x³ × 2/x² = -480x
Thus the coefficient of x in the expansion of (1 + 2/x²)(2x - 3/x)⁵ is: 720-480=240
Regards,
Dragon.Jade :-)
You should use the binomial distribution that states that, for any reals a and b, and any positive integer n:
(a + b)ⁿ = ∑(p=0→n) {n! / [p! × (n - p)!)]} × aⁿ⁻ᵖ × bᵖ
Thus is your case, with a=2x, b=3/x and n=5
(2x - 3/x)⁵ =
5!/(0! × 5!) × (2x)⁵(-3/x)º
+ 5!/(1! × 4!) × (2x)⁴(-3/x)¹
+ 5!/(2! × 3!) × (2x)³(-3/x)²
+ 5!/(3! × 2!) × (2x)²(-3/x)³
+ 5!/(4! × 1!) × (2x)¹(-3/x)⁴
+ 5!/(5! × 0!) × (2x)º(-3/x)⁵
= 1 × 32x⁵
+ 5 × (-48x³)
+ 10 × 72x
+ 10 × (-108/x)
+ 5 × 162/x³
+ 1 × (-243/x⁵)
= 32x⁵ - 240x³ + 720x - 1080/x + 810/x³ - 243/x⁵
Thus the three first terms in descending powers of (2x-3/x)⁵ are:
32x⁵, -240x³ and 720x.
The only two ways to obtain x is:
1 × 720x = 720x and
-240x³ × 2/x² = -480x
Thus the coefficient of x in the expansion of (1 + 2/x²)(2x - 3/x)⁵ is: 720-480=240
Regards,
Dragon.Jade :-)
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(1) By the Binomial Theorem:
(a + b)^n = ∑ [C(n, k) * a^(n - k) * b^k] (from k=0 to n).
So, letting a = 2x, b = -3/x, and n = 5 yields:
(2x - 3/x)^5 = ∑ [C(5, k) * (2x)^(5 - k) * (-3/x)^k] (from k=0 to 5)
= ∑ [C(5, k) * 2^(5 - k) * x^(5 - k) * (-3)^k * x^(-k)] (from k=0 to 5)
= ∑ [C(5, k) * 2^(5 - k) * (-3)^k * x^(5 - 2k)] (from k=0 to 5).
Letting k = 0, 1, and 2 yields the first three terms to be:
(i) Term 1 (k = 0): C(5, 0)*2^(5 - 0)*(-3)^0*x^[5 - 2(0)] = 32x^5
(ii) Term 2 (k = 1): C(5, 1)*2^(5 - 1)*(-3)^1*x^[5 - 2(1)] = -270x^3
(iii) Term k = 3 (k = 2): C(5, 2)*2^(5 - 2)*(-3)^2*x^[5 - 2(2)] = 720x.
(2) Note that, by the distributive property:
(1 + 2/x^2)(2x - 3/x)^5 = (2x - 3/x)^5 + (2/x^2)(2x - 3/x)^5.
(2x - 3/x)^5 and (2/x^2)(2x - 3/x)^5 both produce a x term (720x for (2x - 3/x)^5 and -270x^3/x^2 = -270x for (2/x^2)(2x - 3/x)^5). Therefore, the required x term is:
720x - 270x = 450x,
and the coefficient of x is 450.
I hope this helps!
(a + b)^n = ∑ [C(n, k) * a^(n - k) * b^k] (from k=0 to n).
So, letting a = 2x, b = -3/x, and n = 5 yields:
(2x - 3/x)^5 = ∑ [C(5, k) * (2x)^(5 - k) * (-3/x)^k] (from k=0 to 5)
= ∑ [C(5, k) * 2^(5 - k) * x^(5 - k) * (-3)^k * x^(-k)] (from k=0 to 5)
= ∑ [C(5, k) * 2^(5 - k) * (-3)^k * x^(5 - 2k)] (from k=0 to 5).
Letting k = 0, 1, and 2 yields the first three terms to be:
(i) Term 1 (k = 0): C(5, 0)*2^(5 - 0)*(-3)^0*x^[5 - 2(0)] = 32x^5
(ii) Term 2 (k = 1): C(5, 1)*2^(5 - 1)*(-3)^1*x^[5 - 2(1)] = -270x^3
(iii) Term k = 3 (k = 2): C(5, 2)*2^(5 - 2)*(-3)^2*x^[5 - 2(2)] = 720x.
(2) Note that, by the distributive property:
(1 + 2/x^2)(2x - 3/x)^5 = (2x - 3/x)^5 + (2/x^2)(2x - 3/x)^5.
(2x - 3/x)^5 and (2/x^2)(2x - 3/x)^5 both produce a x term (720x for (2x - 3/x)^5 and -270x^3/x^2 = -270x for (2/x^2)(2x - 3/x)^5). Therefore, the required x term is:
720x - 270x = 450x,
and the coefficient of x is 450.
I hope this helps!