Suma n początkowych wyrazów ciągu arytmetycznego (an) wyraża się wzorem:
Sn = 2 * n^2 +n dla n >/= 1
a) Oblicz sumę 50 poczatkowych wyrazów tego ciągu o numerach parzystych:
a2 + a4 + ... + a100
b) oblicz lim Sn / 3n^2 -2
n->oo
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a)
Sn = 2 n^2 + n; n > = 1
więc
a1 = S1 = 2 + 1 = 3
a1 + a2 = S2 = 2*2^2 + 2 = 2*4 + 2 = 8 + 2 = 10
więc
a2 = 10 -a1 = 10 - 3 = 7
a1+ a2 + a3 = S3 = 2*3^2 + 3 = 18 + 3 = 21
więc
a3 = 21 - ( 3 + 7) = 21 - 10 = 11
r = a2 - a1 = 7 - 3 = 4
Mamy
a1 = 3 i r = 4
Tworzę nowy ciąg bn
b1 = a2 = 7
r1 = 2*r = 2*4 = 8
zatem
b50 = b1 + 49*r1 = 7 + 49*8 = 7 + 392 = 399
oraz
S50 = 0,5 *{ b1 + b50}* 50 = 25* [ 7 + 399 ] = 25 * 406 = 10 150
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b)
cn = [ 2 n^2 + n ] / [ 3 n^2 - 2 ] = [ 2 + 1/n ] / [ 3 - 2/n^2]
więc
lim cn = 2/3
n--> oo
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bo 1/n --> 0 i - 2/n^2 --> 0 , gdyn --> oo
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