Question

Encontre a fórmula fechada de somas e subtrações intercaladas dos n primeiros quadrados.


$\mathsf{S_n=-1^2+2^2-3^2+4^2-5^2+...+(-1)^n\cdot n^2=\displaystyle\sum_{k=1}^n (-1)^k\cdot k^2}$



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Por favor responder de forma detalhada. Respostas com brincadeiras serão eliminadas.

Answer 1

Seja $\mathsf{\Delta(c_{k})=c_{k+1}-c_{k}}$ o operador de diferença anterior de qualquer sequência real $\mathsf{c_{k}}$. Então, para quaisquer $\mathsf{a_{k}}$, $\mathsf{b_{k}}$ sequências e $\mathsf{x}$ constante:

$\bullet\,\,\Delta(a_{k}+ b_{k})=(a_{k+1}+b_{k+1})-(a_{k}-b_{k})=(a_{k+1}-a_{k})+(b_{k+1}-b_{k})\\\\=\Delta(a_{k})+\Delta(b_{k})\\\\\\\bullet\,\,\Delta(x\cdot a_{k})=x\cdot a_{k+1}-x\cdot a_{k}=x\cdot(a_{k+1}-a_{k})=x\cdot\Delta(a_{k})\\\\\\\bullet\,\,\displaystyle\sum_{k=p}^{n}\Delta(a_{k})=\sum_{k=p}^{n}(a_{k+1}-a_{k})=\sum_{k=p}^{n}a_{k+1}-\sum_{k=p}^{n}a_{k}\\\\\\=\sum_{k=p+1}^{n+1}a_{k}-\sum_{k=p}^{n}a_{k}=a_{n+1}+\sum_{k=p+1}^{n}a_{k}-a_{p}-\sum_{k=p+1}^{n}a_{k}=a_{n+1}-a_{p}$
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Vamos encontrar $\mathsf{\Delta(a_{k}),\,\,a_{k}=(-1)^{k}\cdot k^{2}}$:

$\mathsf{\Delta(a_{k})=a_{k+1}-a_{k}}\\\\\mathsf{\Delta(a_{k})=(-1)^{k+1}\cdot(k+1)^{2}-(-1)^{k}\cdot k^{2}}\\\\\mathsf{\Delta(a_{k})=(-1)^{k+1}(k^{2}+2k+1)-(-1)^{k}\cdot k^{2}}\\\\\mathsf{\Delta(a_{k})=(-1)^{k+1}\cdot k^{2}+2(-1)^{k+1}\cdot k+(-1)^{k+1}-(-1)^{k}\cdot k^{2}}\\\\\mathsf{\Delta(a_{k})=(-1)(-1)^{k}\cdot k^{2}+2(-1)(-1)^{k}\cdot k-(-1)^{k}\cdot k^{2}+(-1)(-1)^{k}}\\\\\mathsf{\Delta(a_{k})=-(-1)^{k}\cdot k^{2}-(-1)^{k}\cdot k^{2}-2(-1)^{k}\cdot k-(-1)^{k}}$

$\mathsf{\Delta(a_{k})=-2(-1)^{k}\cdot k^{2}-2(-1)^{k}\cdot k-(-1)^{k}}$

Isolando $\mathsf{(-1)^{k}\cdot k^{2}}$:

$\mathsf{(-1)^{k}\cdot k^{2}=-\dfrac{1}{2}\bigg[\Delta(a_{k})+2(-1)^{k}\cdot k+(-1)^{k}\bigg]}$

Agora vamos achar $\mathsf{\Delta(b_{k}),\,\,b_{k}=(-1)^{k}\cdot k}$:

$\mathsf{\Delta(b_{k})=b_{k+1}-b_{k}}\\\\\mathsf{\Delta(b_{k})=(-1)^{k+1}\cdot(k+1)-(-1)^{k}\cdot k}\\\\\mathsf{\Delta(b_{k})=(-1)^{k+1}\cdot k+(-1)^{k+1}-(-1)^{k}\cdot k}\\\\\mathsf{\Delta(b_{k})=(-1)(-1)^{k}\cdot k-(-1)^{k}\cdot k+(-1)(-1)^{k}}\\\\\mathsf{\Delta(b_{k})=-(-1)^{k}\cdot k-(-1)^{k}\cdot k-(-1)^{k}}\\\\\mathsf{\Delta(b_{k})=-2(-1)^{k}\cdot k-(-1)^{k}}\\\\\Longrightarrow\,\,\mathsf{(-1)^{k}\cdot k=-\dfrac{1}{2}\bigg[\Delta(b_{k})+(-1)^{k}\bigg]}$

Então:

$\mathsf{(-1)^{k}\cdot k^{2}=-\dfrac{1}{2}\bigg[\Delta(a_{k})+2(-1)^{k}\cdot k+(-1)^{k}\bigg]}\\\\\\\mathsf{(-1)^{k}\cdot k^{2}=-\dfrac{1}{2}\bigg[\Delta(a_{k})+2\cdot(-\frac{1}{2})\big[\Delta(b_{k})+(-1)^{k}\big]+(-1)^{k}\bigg]}\\\\\\\mathsf{(-1)^{k}\cdot k^{2}=-\dfrac{1}{2}\bigg[\Delta(a_{k})-\Delta(b_{k})-(-1)^{k}+(-1)^{k}}\bigg]}\\\\\\\mathsf{(-1)^{k}\cdot k^{2}=-\dfrac{1}{2}\Delta(a_{k}-b_{k})}$

Logo,

$\displaystyle\mathsf{\sum_{k=1}^{n}(-1)^{k}\cdot k^{2}=}\\\\\\\mathsf{\sum_{k=1}^{n}-\dfrac{1}{2}\Delta(a_{k}-b_{k})=}\\\\\\\mathsf{-\dfrac{1}{2}\sum_{k=1}^{n}\Delta(a_{k}-b_{k})=}\\\\\\\mathsf{-\dfrac{1}{2}\bigg[\big(a_{n+1}-b_{n+1}\big)-\big(a_{1}-b_{1}\big)\bigg]=}\\\\\\\mathsf{-\dfrac{1}{2}\bigg[(-1)^{n+1}(n+1)^{2}-(-1)^{n+1}(n+1)-(-1)^{1}\cdot1^{2}+(-1)^{1}\cdot1\bigg]=}\\\\\\\mathsf{-\dfrac{1}{2}\bigg[(-1)^{n+1}(n+1)^{2}-(-1)^{n+1}(n+1)\bigg]=}$

$\mathsf{-\dfrac{1}{2}(-1)^{n+1}(n+1)\big[(n+1)-1\big]=}\\\\\\\mathsf{=-\dfrac{1}{2}(-1)(-1)^{n}(n+1)\cdot n}\\\\\\\mathsf{=(-1)^{n}\cdot\dfrac{n(n+1)}{2}}$

Concluímos, portanto, que

$\boxed{\boxed{\mathsf{\sum_{k=1}^{n}(-1)^{k}\cdot k^{2}=(-1)^{n}\cdot\dfrac{n(n+1)}{2}}}}$

Answer 2

Resposta:

Explicação passo a passo:

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