Respuesta :

TheAakash
We'll solve for each one of them:

[tex]\sum_{i=1}^{4}4(5)^{i-1} = (4(5)^{1-1}) + (4(5)^{2-1}) + (4(5)^{3-1}) + (4(5)^{4-1})[/tex]
[tex]=4+20+100+500 = 624 [/tex]

[tex]\sum_{i=1}^{5}3(4)^{i-1} = (3(4)^{1-1}) + (3(4)^{2-1}) + (3(4)^{3-1}) + (3(4)^{4-1}) + (3(4)^{5-1})[/tex]
[tex]=3+12+48+192+768=1023[/tex]

[tex]\sum_{i=1}^{2}5(6)^{i-1} = (5(6)^{1-1}) + (5(6)^{2-1})[/tex]
[tex]=5+30=35[/tex]

[tex]\sum_{i=1}^{4}5^{i-1} = (5^{1-1})+(5^{2-1})+(5^{3-1})+(5^{4-1}) [/tex]
[tex]=1 + 5+ 25+125 = 156[/tex]

So, our totals are: 624, 1023, 35, and 156. So clearly, 1023 > 624 > 156 > 35. W can write it as (your answer):

[tex]\sum_{i=1}^{2}5(6)^{i-1} <\sum_{i=1}^{4}5^{i-1} <\sum_{i=1}^{4}4(5)^{i-1} < \sum_{i=1}^{5}3(4)^{i-1} [/tex]

Hope this helps! If anything is confusing, you can always DM me.
97845

∑ 4 * 5^(i-1) = 4 + 20 + 100 + 500 = 624

∑ 3 * 4^(i-1) = 3 + 12 + 48 + 192 + 768 = 1,023

∑ 5*  6^(i-1) = 5 + 30 = 35

∑ 5^(i-1) = 1 + 5 + 25 + 125 = 156

Answer:

∑ (i=1, 2) 5 * 6^(i-1) < ∑ (i=1, 4) 5^(i-1) < ∑ (i=1, 4) 4 * 5^(i-1) <

< ∑ (i=1, 5) 3 * 4^(i-1)

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