Pascal \u00fc\u00e7geni \u2013 Binom<\/p>\n
Bir k\u00fcmenin alt k\u00fcmelerinin say\u0131s\u0131n\u0131 g\u00f6steren \u201cPASCAL\u201d \u00fc\u00e7genini olu\u015ftural\u0131m.<\/p>\n
K\u00fcmenin Eleman Say\u0131s\u0131:<\/p>\n
<\/p>\n
s(A)=0…………………………………….. ……………1<\/p>\n
s(A)=1…………………………………….. …………1…..1<\/p>\n
s(A)=2…………………………………….. …….1…..2…..1<\/p>\n
s(A)=3…………………………………….. ..1…..3…..3…..1<\/p>\n
s(A)=4……………………………………1. ….4…..6…..4…..1<\/p>\n
s(A)=5………………………………..1….. 5…..10….10…..5….1 …<\/p>\n
<\/p>\n
\u00dc\u00e7genin tepesinde 1 yazd\u0131k.Sonraki sat\u0131rlar\u0131n ilk ve son say\u0131lar\u0131n\u0131 yine 1 ald\u0131k.Bir sat\u0131rda ard\u0131\u015f\u0131k iki say\u0131n\u0131n toplam\u0131n\u0131, bu say\u0131lar\u0131n ortas\u0131na gelecek \u015fekilde bir alt sat\u0131ra yazd\u0131k.Bu i\u015flemlere yukardan a\u015fa\u011f\u0131 do\u011fru devam ettik.<\/p>\n
\u00d6rne\u011fin; s(A)=4 …………..1…..4…..6…..4…..1<\/p>\n
s(A)=5……….1…..5…..10…..10…..5…..1<\/p>\n
Bu tablodaki say\u0131lar\u0131n ne ifade etti\u011fini g\u00f6sterelim.<\/p>\n
A={a,b,c} k\u00fcmesi 3 elemanl\u0131 olup bu k\u00fcmenin alt k\u00fcmelerini yazal\u0131m.<\/p>\n
0 elemanl\u0131 alt k\u00fcmesi{} 1 tane<\/p>\n
1 elemanl\u0131 alt k\u00fcmeleri{a},{b},{c} 3 tane<\/p>\n
2 elemanl\u0131 alt k\u00fcmeleri{a,b},{a,c},{b,c}3 tane<\/p>\n
3 elemanl\u0131 alt k\u00fcmeleri{a,b,c} 1 tane<\/p>\n
<\/p>\n
s(A)=3 olan sat\u0131rdaki say\u0131lar oldu\u011funu g\u00f6r\u00fcn\u00fcz.O halde bu tablo, bir k\u00fcmenin 0 elemanl\u0131, 1 elemanl\u0131, 2 elemanl\u0131,….alt k\u00fcmelerinin say\u0131s\u0131n\u0131 g\u00f6sterir.<\/p>\n
Pascal \u00dc\u00e7genini biraz daha b\u00fcy\u00fcterek a\u015fa\u011f\u0131daki \u00f6rnekleri inceleyelim.<\/p>\n
*6 elemanl\u0131 bir k\u00fcmenin 2 elemanl\u0131 15 tane alt k\u00fcmesi vard\u0131r.(s(A)=6\u2018n\u0131n<\/p>\n
sat\u0131r\u0131ndaki \u00fc\u00e7\u00fcnc\u00fc say\u0131)<\/p>\n
*5 elemanl\u0131 bir k\u00fcmenin 2 elemanl\u0131 en az 3 elemanl\u0131 ka\u00e7 tane alt k\u00fcmesi oldu\u011funu ara\u015ft\u0131ral\u0131m:<\/p>\n
3 elemanl\u0131……….10……….(s(A)=5\u2019in sat\u0131r\u0131nda 4. say\u0131)<\/p>\n
4 elemanl\u0131……….5……….(s(A)=5\u2019in sat\u0131r\u0131nda 5. say\u0131)<\/p>\n
*7 elemanl\u0131 bir k\u00fcmenin en az 2 elemanl\u0131 ka\u00e7 alt k\u00fcmesi oldu\u011funu ara\u015ft\u0131ral\u0131m:<\/p>\n
1.YOL: (21+35+21+7+1)=120<\/p>\n
2.YOL: 2 7-(1+7)=128-8=120 (Neden?)<\/p>\n
<\/p>\n
Binom A\u00e7\u0131l\u0131m\u0131:<\/p>\n
(a+b)n nin a\u00e7\u0131l\u0131m\u0131nda Pascal \u00dc\u00e7genindeki say\u0131lar terimdeki katsay\u0131lar\u0131 olur.a\u2019n\u0131n kuvvetleri n den 0 a kadar azalarak, b\u2019nin kuvvetleri 0 dan n ye kadar artarak yaz\u0131l\u0131r.<\/p>\n
<\/p>\n
<\/p>\n
(a+b)5=?<\/p>\n
Katsay\u0131lar 1 5 10 10 5 1<\/p>\n
A n\u0131n kuvvetleri a5 a4 a3 a2 a 1<\/p>\n
B nin kuvvetleri 1 b b2 b3 b4 b6<\/p>\n
<\/p>\n
(a+b)5=1a5+5a4b+10a3b2+10a2b3+5ab4+1b5<\/p>\n
<\/p>\n
<\/p>\n
*(5x-3y)2=?<\/p>\n
Katsay\u0131lar 1 2 1<\/p>\n
5x\u2019in kuvvetleri 25×2 5x 1<\/p>\n
-3y\u2019nin kuvvetleri 1 -3y 9y2<\/p>\n
(5x-3y)2= 25×2 -2.5x.3y +9y2= 25×2 \u201330xy +9y2<\/p>\n
<\/p>\n
Yukarda ki \u00f6rnekten de g\u00f6r\u00fclebilece\u011fi gibi negatif terimin tek kuvvetlerinin oldu\u011fu terimlerin i\u015fareti negatiftir<\/p>\n","protected":false},"excerpt":{"rendered":"
Pascal \u00fc\u00e7geni \u2013 Binom Bir k\u00fcmenin alt k\u00fcmelerinin say\u0131s\u0131n\u0131 g\u00f6steren \u201cPASCAL\u201d \u00fc\u00e7genini olu\u015ftural\u0131m. K\u00fcmenin Eleman Say\u0131s\u0131: s(A)=0…………………………………….. ……………1 s(A)=1…………………………………….. …………1…..1 s(A)=2…………………………………….. …….1…..2…..1 s(A)=3…………………………………….. ..1…..3…..3…..1 s(A)=4……………………………………1. ….4…..6…..4…..1 s(A)=5………………………………..1….. 5…..10….10…..5….1 … \u00dc\u00e7genin tepesinde 1 yazd\u0131k.Sonraki sat\u0131rlar\u0131n ilk ve son say\u0131lar\u0131n\u0131 yine 1 ald\u0131k.Bir sat\u0131rda ard\u0131\u015f\u0131k iki say\u0131n\u0131n toplam\u0131n\u0131, bu say\u0131lar\u0131n ortas\u0131na gelecek \u015fekilde bir alt … Devam\u0131n\u0131 oku…<\/a><\/p>\n","protected":false},"author":1685,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":[],"categories":[13],"tags":[14299,14298],"_links":{"self":[{"href":"https:\/\/www.sorubak.com\/blog\/wp-json\/wp\/v2\/posts\/40170"}],"collection":[{"href":"https:\/\/www.sorubak.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.sorubak.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.sorubak.com\/blog\/wp-json\/wp\/v2\/users\/1685"}],"replies":[{"embeddable":true,"href":"https:\/\/www.sorubak.com\/blog\/wp-json\/wp\/v2\/comments?post=40170"}],"version-history":[{"count":0,"href":"https:\/\/www.sorubak.com\/blog\/wp-json\/wp\/v2\/posts\/40170\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.sorubak.com\/blog\/wp-json\/wp\/v2\/media?parent=40170"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.sorubak.com\/blog\/wp-json\/wp\/v2\/categories?post=40170"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.sorubak.com\/blog\/wp-json\/wp\/v2\/tags?post=40170"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}