oblicz sumę wszystkich trzycyfrowych liczb naturalnych. Pilne !
a₁=100
an=999
a₂=101
r=101-100=1
an=a₁+(n-1)r
999=100+(n-1)×1
999=100+n-1
n=999-99
n=900
S₉₀₀=½(a₁+a₉₀₀)×900=½(100+999)×900=494 550
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a₁=100
an=999
a₂=101
r=101-100=1
an=a₁+(n-1)r
999=100+(n-1)×1
999=100+n-1
n=999-99
n=900
S₉₀₀=½(a₁+a₉₀₀)×900=½(100+999)×900=494 550