2006-07-19 18:52:10 · 11 answers · asked by .......... 4 in Science & Mathematics ➔ Mathematics
Dear Jimmy,
By my count, the one-thousandth term in the sequence is precisely
535754303593133660474212524530
000905280702405852766803721875
194185175525562468061246599189
407847929063797336458776573412
593572642846157021799228878734
928740196728388741211549271053
730253118557093897709107652323
749179097063369938377958277197
303853145728559823884327108383
021491582631219341860283403468
8.
2006-07-20 01:07:53 · answer #1 · answered by wiseguy 6 · 0⤊ 0⤋
Usually this question is asked about the ones digit....the ones digits of the powers of 2 repeat; 2,4, 8, 16, 32, 64, 128, 256, 512,....So it is 2, 4, 8, 6, over and over again. so 2^999 power, figure out how many groups of 4 in 999...249. 4 * 249 = 996. So 2^996 power will end in 6. So the ones digit is 8. (2^997 ends in 2, 2^998 ends in 4, etc.).
2006-07-20 02:01:34 · answer #2 · answered by Emily C 2 · 0⤊ 0⤋
since the sequence is 2^0, 2^1, 2^2, 2^3... so the 1000th term in the sequence is two raised to the ninety-ninth power, or 633825300114114700748351602688
2006-07-20 02:55:48 · answer #3 · answered by emee_rocks 2 · 0⤊ 0⤋
2 to the 999 power
2006-07-20 01:55:00 · answer #4 · answered by alwaysmoose 7 · 0⤊ 0⤋
For the given sequence...
a, ar , ar², ar³,..................
a = 1
r = 2
a(n) = a * r ^ (n-1)
a(1000) = 1 * 2^999
the 1000th term = 2^999
â 5.3575 * 10^300
2006-07-20 02:57:23 · answer #5 · answered by M. Abuhelwa 5 · 0⤊ 0⤋
the formula is
2^(n - 1)
2^(1000 - 1)
2^(999)
about 5.357543 * 10^300
2006-07-20 11:17:19 · answer #6 · answered by Sherman81 6 · 0⤊ 0⤋
Its a very very big number. Bigger than a googol! A googol is 10^100, but the number you seek is larger than 5.35 x 10^300.
2006-07-20 02:08:05 · answer #7 · answered by TrickMeNicely 4 · 0⤊ 0⤋
2^999
x =2^999
log(x) = 999*log(2)
log(x) ~ 999*0.301029996
log(x) ~ 300.728965668
x ~ 10^(300.7289657)
~ 10^300 *10^(0.728965668)
~ 10^300 *5.357543032
~ 5.357543032*10^300
2006-07-20 23:39:47 · answer #8 · answered by PC_Load_Letter 4 · 0⤊ 0⤋
first term is 2^0
second term is 2^1
third term is 2^2
etc
thousand'th term is 2^999
2006-07-20 01:56:53 · answer #9 · answered by qwert 5 · 0⤊ 0⤋
That's so easy I'm not even going to dignify it with an answer!!
2006-07-20 02:01:08 · answer #10 · answered by Jimmy Pete 5 · 0⤊ 1⤋