Question

Ofer 100 de puncte!!!!

Answer 1

Răspuns:

Explicație pas cu pas:

Answer 2

a)  Avem suma primilor n termeni ai unei progresii geometrice,

cu b₁=1/a² și rația q = 1/a².

Folosim formula:

$\it S_n=b_1\cdot\dfrac{q^n-1}{q-1}$

Vom avea:

$\it S_n=\dfrac{1}{a^2}\cdot\dfrac{\Big(\dfrac{1}{a^2}\Big)^n-1}{\dfrac{1}{a^2}-1}=\dfrac{1}{a^2}\cdot\dfrac{\dfrac{1}{a^{2n}}-1}{\dfrac{1}{a^2}-1}=\dfrac{1}{a^2}\cdot\dfrac{\dfrac{1-a^{2n}}{a^{2n}}}{\dfrac{1-a^2}{a^2}}=\\ \\ \\ =\dfrac{1}{\not a^2}\cdot\dfrac{1-a^{2n}}{a^{2n}}\cdot\dfrac{\not a^2}{1-a^2}=\dfrac{a^{2n}-1}{a^{2n}(a^2-1)}$

$\it b)\ \ (\dfrac{1}{a}+a)^2+(\dfrac{1}{a^2}+a^2)^2= \dfrac{1}{a^2}+2+a^2+\dfrac{1}{a^4}+2+a^4$

Din desfășurarea primilor doi termeni ai sumei, se observă că avem:

$\it S= \underbrace{\it2+2+2+\ ...\ +2}_{n\ termeni}+(a^2+a^4+\ ...\ +a^{2n})+\Big(\dfrac{1}{a^2}+\dfrac{1}{a^4}+\ ...\ +\dfrac{1}{a^{2n}}\Big)=\\ \\ \\ =2n+a^2\cdot\dfrac{a^{2n}-1}{a^2-1}+\dfrac{a^{2n}-1}{a^{2n}(a^2-1)}=2n+\dfrac{a^{2n}-1}{a^2-1}\Big(\dfrac{1}{a^{2n}}+a^2\Big)=\\ \\ \\ =2n+\dfrac{a^{2n}-1}{a^2-1}\cdot\dfrac{1+a^{2n+2}}{a^{2n}}$