標題:
發問:
13. Show that the coefficient of x^n in the expansion of (2x+1)e^x is (2n+1)/n!. 14 Find the coefficient of x^6 in the expansion of (e^x + 2)^2 15.Expand (e^x - e^-x)/2 in ascending powers of x up to the term x^5.By putting x=1, show that e - 1/e => 47/20
最佳解答:
ex = ∑(k = 0 to infinity) xk / k! 13. (2x + 1)ex = 2xex + ex xn term in the first part is (2)[1/(n – 1)!] xn term in the second part is 1/n! So xn term is 2/(n – 1)! + 1/n! = 2n/n! + 1/n! = (2n + 1)/n! 14 (ex + 2)2 = e2x + 4ex + 4 x6 term in first expression is (2)6/6! = 64/720 x6 term in second expression is (4)(1/6!) = 4/720 x6 term is 64/720 + 4/720 = 17/180 15.(ex – e-x)/2 = [(1 + x + x2/2! + x3/3! + x4/4! + x5/5! + …) – (1 – x + x2/2! – x3/3! + x4/4! – x5/5! + …)]/2 = x + x3/3! + x5/5! Putting x = 1, (e – 1/e)/2 → 1 + 1/3! + 1/5! = 1 + 1/6 + 1/120 = 141 / 120 = 47/40 e – 1/e → 47/20
其他解答:
I don't understant the first question why the first part in (n-1)! and second part is n!
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Expotential series question發問:
13. Show that the coefficient of x^n in the expansion of (2x+1)e^x is (2n+1)/n!. 14 Find the coefficient of x^6 in the expansion of (e^x + 2)^2 15.Expand (e^x - e^-x)/2 in ascending powers of x up to the term x^5.By putting x=1, show that e - 1/e => 47/20
最佳解答:
ex = ∑(k = 0 to infinity) xk / k! 13. (2x + 1)ex = 2xex + ex xn term in the first part is (2)[1/(n – 1)!] xn term in the second part is 1/n! So xn term is 2/(n – 1)! + 1/n! = 2n/n! + 1/n! = (2n + 1)/n! 14 (ex + 2)2 = e2x + 4ex + 4 x6 term in first expression is (2)6/6! = 64/720 x6 term in second expression is (4)(1/6!) = 4/720 x6 term is 64/720 + 4/720 = 17/180 15.(ex – e-x)/2 = [(1 + x + x2/2! + x3/3! + x4/4! + x5/5! + …) – (1 – x + x2/2! – x3/3! + x4/4! – x5/5! + …)]/2 = x + x3/3! + x5/5! Putting x = 1, (e – 1/e)/2 → 1 + 1/3! + 1/5! = 1 + 1/6 + 1/120 = 141 / 120 = 47/40 e – 1/e → 47/20
其他解答:
I don't understant the first question why the first part in (n-1)! and second part is n!
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