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1, 227 và 318
Ta có:
227 = (23)9 = 89
318 = (32)9 = 99
Vì 8 < 9 nên 89 < 99 hay 227 < 318
2, 2225 và 350
Ta có:
2225 = (29)25 = 51225
350 = (32)25 = 925
Vì 512 > 9 nên 51225 > 925 hay 2225 > 350
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Bài 1 : a, Ta có : (-1)3 . (-1)5 . (-1)7 . (-1)9 . (-1)11 . (-1)13
= (-1)(-1).(-1).(-1).(-1).(-1)
= (-1)6
= 1
b, (1000 - 13) . (1000 - 23) . (1000 - 33) . ... . (1000 - 503)
= (1000 - 13) . (1000 - 23) . (1000 - 33) .... (1000 - 103).......(1000 - 503)
= (1000 - 13) . (1000 - 23) . (1000 - 33) .... 0 ........(1000 - 503)
= 0
Bài 2 :
Đặt A = 12 + 22 + 32 + ... + 102 = 385
=> 22(12 + 22 + 32 + ... + 102) = 22.385
=> 22 + 42 + 62 + ..... + 202 = 4.385
=> 22 + 42 + 62 + ..... + 202 = 1540
Vậy 22 + 42 + 62 + ..... + 202 = 1540
bài 3:
a) 2S=2+22+23+24+...+251
2S-S=251-1
mà 251-1<251
Suy ra:s<251
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a) có 231=2.230=2.810
321=3.320=3.910
vì 2.810 < 3.910 nên 231 < 321
b)
có S = 1 + 2 + ... + 250
<=> S = 20 + 21 + 22 + 23 + ... + 250
=> 2S = 2(20 + 21 + 22 + 23 + ... + 250) = 21 + 22 + 23 + ... + 251
=> 2S - S = 21 + 22 + 23 + ... + 251 - ( 20 + 21 + 22 + 23 + ... + 250)
=> S = 21 + 22 + 23 + ... + 251 - 20 - 21 - 22 - 23 - ... - 250
=> S = 251 - 20
=> S = 251 -1 < 251
=> S < 251
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\(B=\frac{-1}{3}+\frac{1}{3^2}-\frac{1}{3^3}+\frac{1}{3^4}-...+\frac{1}{3^{50}}-\frac{1}{3^{51}}\)
\(3B=-1+\frac{1}{3}-\frac{1}{3^2}+\frac{1}{3^3}-...+\frac{1}{3^{49}}-\frac{1}{3^{50}}\)
\(3B+B=\left(-1+\frac{1}{3}-\frac{1}{3^2}+\frac{1}{3^3}-...+\frac{1}{3^{49}}-\frac{1}{3^{50}}\right)+\left(\frac{-1}{3}+\frac{1}{3^2}-\frac{1}{3^3}+\frac{1}{3^4}-...+\frac{1}{3^{50}}-\frac{1}{3^{51}}\right)\)
\(4B=-1-\frac{1}{3^{51}}\)
\(B=\frac{-1-\frac{1}{3^{51}}}{4}\)
S= 1+3+32+.......+350 (1)
=> 3S= 3+32+33+......+351 (2)
Lấy (2)-(1) ta có:
3S-S= (3+32+33+..........+351)-(1+3+32+..................+350)
2S= 351-1
S= \(\frac{3^{51}-1}{2}\)