Question

Sa se calculeze C de 16 luate cate 0 + C de 16 luate cate 2 + C de 16 luate cate 4 + ... + C de 16 luate cate 16.

Answer 1

$(1+1)^n=C_n^0+C_n^1+C_n^2\ldots+C_n^n;\\(1-1)^n=C_n^0-C_n^1+C_n^2\ldots+C_n^n.\\Dac\breve{a}\ le\ adun\breve{a}m\;membru\ cu\ membru,\ ob\c{t}inem:\\2^n=2\cdot(C_n^0+C_n^2+\ldots),\ deci\;C_n^0+C_n^2+\ldots=2^{n-1}.\\Pentru\;n=16\;te\;las\;pe\ tine\ s\breve{a}\;calculezi.\;Spor\;la\;treab\breve{a}.\\\\Green\;eyes.$

Answer 2

Binomul lui Newton este definit ca 
$(x+y)^{n}=\sum_{k=0}^{n}{C_{n}^{k}x^{n-k}y^{k}}$
Iar pentru cazul in care sunt scazute
$(x-y)^{n}=\sum_{k=0}^{n}{C_{n}^{k}(-1)^{k}x^{n-k}y^{k}}$
Daca inlocuim pe x si y cu 1 avem
$(1+1)^{n}=\sum_{k=0}^{n}{C_{n}^{k}1^{n-k}1^{k}}=\sum_{k=0}^{n}{C_{n}^{k}=C_{n}^{0}+C_{n}^{1}+C_{n}^{2}+..+C_{n}^{n-1}+C_{n}^{n}=2^{n}$
Iar in cazul In care inlocuim in a doua ecuatie cu 1 si 1 si presupunem ca n=2p
$(1-1)^{n}=\sum_{k=0}^{n}{C_{n}^{k}(-1)^{k}1^{n-k}1^{k}}=C_{n}^{0}-C_{n}^{1}+C_{n}^{2}-C_{n}^{3}+...-C_{n}^{2p-1}+C_{n}^{2p}=0$
Deci observi ca la scadere atunci cand ai k impar, atunci (-1)^k devine negativ, deci termenii impari sunt scazuti. Deci daca aduni cele 2 forme ale binomului, termenii pari se dubleaza si cei impari se reduc unul cu altul
$(1+1)^{n}+(1-1)^{n}=2C_{n}^{0}+2C_{n}^{2}+2C_{n}^{4}+...+2C_{n}^{n}=2^{n}+0=2^{n}\Rightarrow C_{n}^{0}+C_{n}^{2}+C_{n}^{4}+...+C_{n}^{n}=2^{n-1}$
Deci in cazul nostru
$C_{16}^{0}+C_{16}^{2}+C_{16}^{4}+...+<span>C_{16}^{16}=2^{16-1}=2^{15}</span>$