Question

DacI (aₓ)ₓ≥1 este o progresie aritmetica de ratie r, atunci sa se calculeze
suma:

S=a₁a₂+a₃a₄+...+aₓ₋₁aₓ

Answer 1

Vom scrie suma asa:

$S=\sum_{k=1}^{x/2} a_{2k-1}a_{2k}$

Formula generala a termenului unei progresii aritmetice:
$a_n=a_1+(n-1)r$

Inlocuim in suma:
[tex]S=\sum_{k=1}^{x/2} (a_1+(2k-2)r)(a_1+(2k-1)r)\\\\ S=\sum_{k=1}^{x/2}(a_1^2+(4k-3)a_1r+(2k-1)(2k-2)r^2)[/tex]

Impartim suma noastra in 3 sume mai mici:

[tex]s_1=\sum_{k=1}^{x/2}a_1^2\\ s_2=\sum_{k=1}^{x/2}(4k-3)a_1r\\ s_3=\sum_{k=1}^{x/2}(2k-1)(2k-2)r^2[/tex]

$s_1= \frac{x}{2}\cdot a_1^2$

[tex]s_2=a_1r\sum_{k=1}^{x/2}(4k-3)=a_1r(\sum_{k=1}^{x/2}2k-\sum_{k=1}^{x/2}3)\\ s_2=a_1r(2 \frac{(x/2)(x/2+1)}{2}-3*x/2 )\\ s_2=a_1r \cdot \frac{x}{2}( \frac{x}{2}-2) \\ s_2=a_1r\cdot \frac{x(x-4)}{4} [/tex]

[tex]s_3=r^2\sum_{k=1}^{x/2}(2k-1)(2k-2)=r^2\sum_{k=1}^{x/2}(4k^2-6k+2)\\ s_3=r^2(\sum_{k=1}^{x/2}4k^2-\sum_{k=1}^{x/2}6k+ \sum_{k=1}^{x/2}2)\\ s_3=r^2(4\frac{(x/2)(x/2+1)(x+1)}{6} -6 \frac{(x/2)(x/2+1)}{2} +x)\\ s_3=r^2(( \frac{x}{2} )( \frac{x}{2} +1)( \frac{2(x+1)}{3}-3 )+x )\\ s_3=r^2( \frac{x}{2}\cdot \frac{x+2}{2} \cdot \frac{2x-7}{3} +x)\\\\ s_3=r^2( \frac{x(x+2)(2x-7)}{12} +x)[/tex]

[tex]S=s_1+s_2+s_3= a_1^2\cdot \frac{x}{2}+a_1r\cdot \frac{x(x-4)}{4}+r^2( \frac{x(x+2)(2x-7)}{12} +x)\\ [/tex]