Câu hỏi của Nguyễn Phương Ngân

Question

Tính nhanha)S1=5+52+53+......+599+5100b)S2=2+22+23+...+299+2100

Lê Chí Cường · Accepted Answer

a)Ta có:S1=5+52+53+…+599+5100=>5.S1=52+53+54+…+5100+5101=>5.S1-S1=52+53+54+…+5100+5101-5-52-53-…-599-5100=>4.S1=5101-5=>$S_1=\frac{5^{101}-5}{4}$b)S2=2+22+23+…+299+2100=>2.S2=22+23+24+…+2100+2101=>2.S2-S2=22+23+24+…+2100+2101-2-22-23-…-299-2100=>S2=2101-2Câu trả lời: a)Ta có:S1=5+52+53+…+599+5100=>5.S1=52+53+54+…+5100+5101=>5.S1-S1=52+53+54+…+5100+5101-5-52-53-…-599-5100=>4.S1=5101-5=>$S_1=\frac{5^{101}-5}{4}$b)S2=2+22+23+…+299+2100=>2.S2=22+23+24+…+2100+2101=>2.S2-S2=22+23+24+…+2100+2101-2-22-23-…-299-2100=>S2=2101-2

Ngô Huy Hoàng · Answer

2S1=52+53+54+....+5100+51012S1-s1=5101-5S1=5101-5b) S2=2101-2Câu trả lời: 2S1=52+53+54+....+5100+51012S1-s1=5101-5S1=5101-5b) S2=2101-2

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