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Pregunta: Si $S_1,S_2,S_3,...,S_n$ ser la suma de $n$ infinito G. P. de la serie, respectivamente, cuyos primeros términos son, respectivamente, $1,2,3,...,n$ y coeficientes comunes $\frac{1}{2},\frac{1}{3},\frac{1}{4},...,\frac{1}{n+1}$ respectivamente, y luego encontrar el valor de $S_1^2+S_2^2+S_3^2+...+S_{2n-1}^2$.

Mi intento:

$S_k=\frac{k}{1-\frac{1}{k+1}}$

$\implies S_k=k+1$

$\implies S_k^2=(k+1)^2$

$\implies S=\sum\limits_{k=1}^{2n-1}S_k^2$

$\implies S=\sum\limits_{k=1}^{2n-1}(k+1)^2$

$\implies S=\sum\limits_{k=1}^{2n-1}(k^2+1+2k)$

$\implies S=\sum\limits_{k=1}^{2n-1}(k^2)+\sum\limits_{k=1}^{2n-1}(1)+\sum\limits_{k=1}^{2n-1}(2k)$

$\implies S=\frac{(2n-1)(2n)[2(2n-1)+1]}{6}\ \ \ \ \ \ \ +(2n-1)+\frac{2(2n-1)(2n)}{2}$

$\implies S=\frac{(4n^2-2n)(4n-1)}{6}\;\;\;+2n-1+4n^2-2n$

$\implies S=\frac{16n^3-4n^2-8n^2+2n}{6}\;\;+2n-1+4n^2-2n$

$\implies S=\frac{16n^3-4n^2-8n^2+2n-6+24n^2}{6}$

$\implies S=\frac{16n^3+12n^2+2n-6}{6}$

$\implies S=\frac{8n^3+6n^2+n-3}{3}$

Mi problema: La respuesta correcta es $\frac{n(2n+1)(4n+1)}{3}$ que en la simplificación da $\frac{8n^3+6n^2+n}{3}$. Donde he ido mal? Por favor, ayudar.

Answer 1

$$ \begin{align} \sum_{r=1}^{2n-1}S_r^2&=\sum_{r=1}^{2n-1}\left(\frac r{1-\frac 1{r+1}}\right)^2\\ &=\sum_{r=1}^{2n-1}(r+1)^2\\ &=\sum_{r=2}^{2n}r^2\\ &=\frac 16 (2n)(2n+1)(4n+1)-1\\ &=\boxed{\frac {n(2n+1)(4n+1)}3}-1\\ &=\frac {8n^3+6n^2+n-3}3\\ &=\frac {(2n-1)(4n^2+5n+3)}3\end {align} $$

Answer 2

Ya que $$ \begin{align} (k+1)^2 &=(k+1)k+(k+1)\\ &=2\binom{k+1}{2}+\binom{k+1}{1} \end {align} $$ tenemos $$ \begin{align} \sum_k\binom{2n-1-k}{2n-1-k}\left[2\binom{k+1}{k-1}+\binom{k+1}{k}\right] &=2\binom{2n+1}{2n-2}+\binom{2n+1}{2n-1}\\ &=\frac{8n^3+6n^2+n}3 \end {align} $$ Por lo tanto, al omitir el término$k=0$, obtenemos $$ \ sum_ {k = 1} ^ {2n-1} (k +1) ^ 2 = \ bbox [5px, borde: 2px sólido # C0A000] {\ frac {8n ^ 3 +6n ^ 2 + n-3} 3} $$