Question

Sendo Xn a soma dos n primeiros termos da sequência (3,5,7,9,11...) , e Yn o n- ésimo termo da sequência (-3,-35,-67,-99...) então, asoma dos valores de n sabendo que Xn = |Yn|igual a:A 32. C 29.B 34. D 30.

Answer 1

$\mathrm{\Rightarrow S_1=\{3,5,7,9,11,\dots\}\ \to\ r=2\ \ \|\ \ n=n}\\\\ \mathrm{a_n=a_1+(n-1)r\ \to\ a_n=3+(n-1)2\ \to\ a_n=2n+1}\\\\ \mathrm{X_n=n\bigg(\dfrac{a_1+a_n}{2}\bigg)\ \to\ X_n=n\bigg(\dfrac{3+2n+1}{2}\bigg)\ \to\ \boxed{\mathrm{X_n=n(n+2)}}}$

$\mathrm{\Rightarrow S_2=\{-3,-35,-67,-99,\dots\}\ \to\ r=-32\ \ \|\ \ n=n}\\\\ \mathrm{Y_n=a_1+(n-1)r\ \to\ Y_n=-3+(n-1)(-32)\ \to\ \boxed{\mathrm{Y_n=29-32n}}}$

$\mathrm{\Rightarrow X_n=|Y_n|\ \to\ n(n+2)=|29-32n|\ \to\ n(n+2)=32n-29}\\\\ \mathrm{n^2+2n=32n-29\ \to\ n^2-30n+29=0}\\\\ \mathrm{n=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}=\dfrac{-(-30)\pm\sqrt{(-30)^2-4.1.29}}{2.1}=}\\\\ \mathrm{=\dfrac{30\pm\sqrt{900-116}}{2}=\dfrac{30\pm\sqrt{784}}{2}=\dfrac{30\pm28}{2}=15\pm14}\\\\ \mathrm{*\ n=15+14\ \to\ \boxed{\mathrm{n=29}}\ \ \big\|\ *\ n=15-14\ \to\ \boxed{\mathrm{n=1}}}$

$\mathrm{\Rightarrow Soma\ dos\ valores\ de\ n\ \to\ 29+1=\boxed{\boxed{\mathbf{30}}}}$