Question

Problema 25. Punctele a, b.

Answer 1

Răspuns

Explicație pas cu pas:

$a) a_n=a_1+(n-1)\cdot r \Rightarrow \dfrac{1}{m}=a_1+(n-1)\cdot r \\ a_m=a_1+(m-1)\cdot r \Rightarrow \dfrac{1}{n}=a_1+(m-1)\cdot r\\Scadem~cele~doua~relatii :  \dfrac{1}{m}-\dfrac{1}{n}=(n-1)\cdot r -(m-1)\cdot r\\\dfrac{n-m}{mn}=r(n-1-m+1)\\\dfrac{n-m}{mn}=r(n-m) |:(n-m) ,~putem~imparti~deoarece~m\neq n\\\boxed{r=\dfrac{1}{mn}}\\Daca~stim~ratia~putem~calcula~si~primul~termen:\\\dfrac{1}{n}=a_1+ \dfrac{m-1}{mn}\\a_1=\dfrac{1}{n}-\dfrac{m-1}{mn}=\dfrac{m-m+1}{mn}= \dfrac{1}{mn}$

$a_{m\cdot n}=a_1+(m\cdot n-1)\cdot r=\dfrac{1}{mn}+\dfrac{mn-1}{mn}=\dfrac{1}{mn}+1-\dfrac{1}{mn}=\boxed{1}\\\\S_{m\cdot n}=\dfrac{(a_1+a_{m\cdot n})\cdot mn}{2}=\dfrac{\left(\frac{1}{mn}+1\right)\cdot mn}{2}=\boxed{\dfrac{mn+1}{2}}$

$b)S_{10}=a_{10}+S_9\\\dfrac{3\cdot 10^2-10}{2}=a_{10}+\dfrac{3\cdot 9^2-9}{2}\\\\\dfrac{3\cdot 100-10}{2}-\dfrac{3\cdot 81-9}{2}=a_{10}\\\\\dfrac{300-10}{2}-\dfrac{243-9}{2}=a_{10}\\\\\dfrac{290}{2}-\dfrac{234}{2}=a_{10}\\145-117=a_{10}\\\boxed{A=a_{10}=28}$

$a_n=S_n-S_{n-1}\\a_n=\dfrac{3n^2-n}{2}-\dfrac{3(n-1)^2-(n-1)}{2}= \\a_n= \dfrac{3n^2-n-3(n^2-2n+1)+n-1}{2} \\ a_n=\dfrac{3n^2-n-3n^2+6n-3+n-1}{2}\\a_n= \dfrac{6n-4}{2}=\boxed{3n-2}$

$\displaystyle T=a_1+a_3+a_5+\ldots+a_{2n-1}\\T=\sum_{k=1}^n a_{2k-1} =\sum_{k=1}^n (3 (2k-1)-2) = \sum_{k=1}^n (6k-3-2)=\\=\sum_{k=1}^n (6k-5) =6\sum_{k=1}^n k-5n=6\cdot \dfrac{n(n+1)}{2}-5n=\\=3n(n+1)-5n=3n^2+3n-5n=3n^2-2n=a_n~:)$