Question

Folosind binomul lui Newton si folosind formula lui Tk+1 sa se calculeze T6 :
a) (√5 +1)⁵
b)(√2-2)⁷
c) (√3-∛3)⁹
d)(x+∛x)¹³ , x>0

Answer 1

formula $T_{k+1}=C^{k} _{n}*a^{n-k}*b^{k}$  . k va fi 5 in toate ex daca iti cere T₆,fiindca este egal cu T₅₊₁
a) a=√5=$5^{ \frac{1}{2} }$
b=1
n=5
T₆=T₅₊₁=$C^{5} _{5}*(5^{ \frac{1}{2} })^{5-5} *1^{5} =C^{5} _{5}* 1*1=C^{5} _{5} =1$
b)a=√2=$2^{ \frac{1}{2} }$
b=-2
n=7
T₅₊₁=$C^{5} _{7}*(2 ^{ \frac{1}{2} })^{7-5} *(-2) ^{5}= \frac{7!}{2!*5!} *2*(-2)^{5} =21*2*(-32)=-1344$
c)a=√3=$3^{ \frac{1}{2} }$
b=-∛3=$-3^{ \frac{1}{3} }$
n=9
T₅₊₁=$C^{5} _{9} *(3^{ \frac{1}{2} })^{9-5}*((-3)^{ \frac{1}{3} })^{5}=- \frac{9!}{4!*5!}*3^{ \frac{11}{3} } =-126*3^{ \frac{11}{3} }$
d)T₅₊₁=$C^{5} _{13}*x^{13-5}*(x^{ \frac{1}{3} })^{5} = \frac{13!}{8!*5!}* x^{ \frac{29}{3} } =1287*x^{ \frac{29}{3} }$