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linda_nm

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[ÇóÖú] Çë½ÌÊýѧÐòÁÐÇóºÍÎÊÌâ! ÒÑÓÐ3È˲ÎÓë

Çë½ÌÒ»¸öÐòÁÐ(1)^2,(1+a)^2,(1+a+a^2)^2, ..., (+a+a^2+a^3+..+a^k)^2µÄºÍʽ×Ó

[ Last edited by feixiaolin on 2014-12-21 at 12:05 ]
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wurongjun

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feixiaolin: ½ð±Ò+4 2014-12-21 14:51:57
>> syms k
>> symsum(a^k,k,0,k)
ans =
a^(k+1)/(a-1)-1/(a-1)
>> factor(ans)
ans =
(a^(k+1)-1)/(a-1)
>> symsum((a^(k+1)-1)/(a-1),k,0,k)
ans =
-(-a^(k+1)*a+(k+1)*a-k-1)/(a-1)^2-a/(a-1)^2
>> factor(ans)
ans =
-(-a^(k+1)*a+k*a+2*a-k-1)/(a-1)^2
ÉÆ¶ñµ½Í·ÖÕÓб¨,È˼äÕýµÀÊDz×É£.
4Â¥2014-12-21 12:12:55
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pippi6

Ìú¸Ëľ³æ (ÖøÃûдÊÖ)

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¡¾´ð°¸¡¿Ó¦Öú»ØÌû

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feixiaolin: ½ð±Ò+4 2014-12-21 14:52:07
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Çë½ÌÊýѧÐòÁÐÇóºÍÎÊÌâ!
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9Â¥2014-12-21 14:31:23
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zaq123321

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linda_nm(feixiaolin´ú·¢): ½ð±Ò+5 2014-12-21 14:51:48
Can you double check whether the third item is (1+a+a^2)^2 or not? If that's true, it will be the sum of squared geometric series. Should be not complicated. Good luck.

[ ·¢×ÔÊÖ»ú°æ http://muchong.com/3g ]
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2Â¥2014-12-21 10:28:01
ÒÑÔÄ   »Ø¸´´ËÂ¥   ¹Ø×¢TA ¸øTA·¢ÏûÏ¢ ËÍTAºì»¨ TAµÄ»ØÌû

feixiaolin

ÈÙÓþ°æÖ÷ (ÎÄ̳¾«Ó¢)

ÓÅÐã°æÖ÷

ÿһÏî*(1-a)^2/(1-a)^2£¬¼ò»¯·Ö×Ó±í´ïʽ£¬ÔÙÏà¼Ó

» ±¾ÌûÒÑ»ñµÃµÄºì»¨£¨×îÐÂ10¶ä£©

3Â¥2014-12-21 12:06:12
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linda_nm

гæ (³õÈëÎÄ̳)

ÒýÓûØÌû:
2Â¥: Originally posted by zaq123321 at 2014-12-21 10:28:01
Can you double check whether the third item is (1+a+a^2)^2 or not? If that's true, it will be the sum of squared geometric series. Should be not complicated. Good luck.
...

ÕýÊÇ(1+a+a^2)^2
5Â¥2014-12-21 13:41:17
ÒÑÔÄ   »Ø¸´´ËÂ¥   ¹Ø×¢TA ¸øTA·¢ÏûÏ¢ ËÍTAºì»¨ TAµÄ»ØÌû

linda_nm

гæ (³õÈëÎÄ̳)

ÒýÓûØÌû:
4Â¥: Originally posted by wurongjun at 2014-12-21 12:12:55
>> syms k
>> symsum(a^k,k,0,k)
ans =
a^(k+1)/(a-1)-1/(a-1)
>> factor(ans)
ans =
(a^(k+1)-1)/(a-1)
>> symsum((a^(k+1)-1)/(a-1),k,0,k)
ans =
-(-a^(k+1)*a+(k+1)*a-k-1)/( ...

ÇëÎÊÕâ¸öÊÇʲô³ÌÐò°¡?ÎÒ¿´²»¶®°¡
6Â¥2014-12-21 13:44:56
ÒÑÔÄ   »Ø¸´´ËÂ¥   ¹Ø×¢TA ¸øTA·¢ÏûÏ¢ ËÍTAºì»¨ TAµÄ»ØÌû

linda_nm

гæ (³õÈëÎÄ̳)

ÒýÓûØÌû:
2Â¥: Originally posted by zaq123321 at 2014-12-21 10:28:01
Can you double check whether the third item is (1+a+a^2)^2 or not? If that's true, it will be the sum of squared geometric series. Should be not complicated. Good luck.
...

ÇëÎÊÄÜÏêϸµÄдһÏÂÂð?
7Â¥2014-12-21 13:46:23
ÒÑÔÄ   »Ø¸´´ËÂ¥   ¹Ø×¢TA ¸øTA·¢ÏûÏ¢ ËÍTAºì»¨ TAµÄ»ØÌû

linda_nm

гæ (³õÈëÎÄ̳)

Ëͺ컨һ¶ä
ÒýÓûØÌû:
3Â¥: Originally posted by feixiaolin at 2014-12-21 12:06:12
ÿһÏî*(1-a)^2/(1-a)^2£¬¼ò»¯·Ö×Ó±í´ïʽ£¬ÔÙÏà¼Ó

¿ÉÒÔ¸øÒ»Ï¾ßÌå¹ý³ÌÂð?
8Â¥2014-12-21 13:49:06
ÒÑÔÄ   »Ø¸´´ËÂ¥   ¹Ø×¢TA ¸øTA·¢ÏûÏ¢ ËÍTAºì»¨ TAµÄ»ØÌû

feixiaolin

ÈÙÓþ°æÖ÷ (ÎÄ̳¾«Ó¢)


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10Â¥2014-12-21 14:53:13
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