Question

Să se calculeze primii 3 termeni ai șirurilor de numere reale date prin formula termenului general astfel:
a) an= 2n-3 , n>=1
b) bn = n-2 supra n+1 , n>=1
c) xn = sin npi supra 6 , n>= 1
d) un = 1 supra 1×2 + 1 supra 2×3 +...+ 1 supra n(n+1)

Answer 1

 

$\displaystyle\bf\\a)~~~a_n=2n-3,~~n\geq1\\a_1=2\cdot1-3=2-3=-1\\a_2=2\cdot2-3=4-3=1\\a_3=2\cdot3-3=6-3=3\\--------\\\\b)~~~b_n=\frac{n-2}{n+1},~~n\geq1\\\\b_1=\frac{1-2}{1+1}=-\frac{1}{2}\\\\b_2=\frac{2-2}{2+1}=0\\\\b_3=\frac{3-2}{3+1}=\frac{1}{4}\\\\------$

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$\displaystyle\bf\\c)~~~x_n=sin\,\frac{n\pi}{6},~~n\geq1\\\\x_1=sin\,\frac{1\pi}{6}=sin\,\frac{1\pi}{6}=sin\,30^o=\frac{1}{2}\\\\x_2=sin\,\frac{2\pi}{6}=sin\,\frac{\pi}{3}=sin\,60^o=\frac{\sqrt{3}}{2}\\\\x_3=sin\,\frac{3\pi}{6}=sin\,\frac{\pi}{2}=sin\,90^o=1\\\\-------$

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$\displaystyle\bf\\d)~~~u_n=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\cdots+\frac{1}{n(n+1)},~~n\geq1\\\\u_1=\frac{1}{1\cdot2}=\frac{1}{2}\\\\u_2=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}=\frac{1}{1}-\frac{1}{3}=\frac{2}{3}\\\\u_3=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}=\frac{1}{1}-\frac{1}{4}=\frac{3}{4}$