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(a) Compute the sum

 

\(101^2 - 97^2 + 93^2 - 89^2 + \cdots + 5^2 - 1^2.\)


(b) Compute the sum

 

\((a +(2n+1)d)^2- (a + (2n)d)^2 +(a + (2n-1)d)^2 - (a+(2n-2)d)^2 + \cdots + (a+d)^2 - a^2.\)

 Jul 1, 2020
 #1
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See both answers by Alan on this link: https://web2.0calc.com/members/alan/?answerpage=751

See the answer to the second question by Max Wong here: https://web2.0calc.com/questions/please-help-question-not-answered#r2

 Jul 1, 2020
 #2
avatar+25541 
+3

(a) Compute the sum:
\(101^2-97^2+93^2-89^2+\ldots+5^2-1^2\)


\(\begin{array}{|rcll|} \hline && \mathbf{ 101^2-97^2+93^2-89^2+\ldots+5^2-1^2} \\\\ &=& 101^2 + 93^2 + 85^2 +\ldots+ 13^2+5^2 \\ && -97^2 -89^2 -81^2 -\ldots- 9^2-1^2 \\\\ &=& \mathbf{\sum \limits_{k=1}^{13}\Big(~ 8k-3~\Big)^2 -\sum \limits_{k=1}^{13}\Big(~ 8k-7~\Big)^2} \\ \hline \end{array}\)

 

\(\begin{array}{|rcll|} \hline && \mathbf{\sum \limits_{k=1}^{13}\Big(~ 8k-3~\Big)^2 -\sum \limits_{k=1}^{13}\Big(~ 8k-7~\Big)^2} \\\\ &=& \sum \limits_{k=1}^{13}\Big(~ 8k-3~\Big)^2 - \Big(~ 8k-7~\Big)^2 \\\\ &=& \sum \limits_{k=1}^{13}64k^2-48k+9-(64k^2-112k+49) \\\\ &=& \sum \limits_{k=1}^{13} {\color{red}64k^2}-48k+9{\color{red}-64k^2}+112k-49 \\\\ &=& \sum \limits_{k=1}^{13} -48k+9+112k-49 \\\\ &=& \sum \limits_{k=1}^{13} 64k-40 \\\\ &=& 64\sum \limits_{k=1}^{13}k -\sum \limits_{k=1}^{13}40 \quad | \quad \sum \limits_{k=1}^{13}40 = 40*13 \\\\ &=& 64\sum \limits_{k=1}^{13}k - 40*13 \quad | \quad \sum \limits_{k=1}^{13}k = \dfrac{(1+13)}{2}*13 \\\\ &=& 64\dfrac{(1+13)}{2}*13 - 40*13 \\\\ &=& 32*14*13 - 40*13 \\\\ &=& 5824 - 520 \\\\ &=& \mathbf{5304} \\ \hline \end{array}\)

 

\(\mathbf{ 101^2-97^2+93^2-89^2+\ldots+5^2-1^2} = \mathbf{5304}\)

 

laugh

 Jul 1, 2020
 #4
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0

(a)  listfor(n, 1, 26,  (-1)^(n + 1) *(105 - 4 *n)^2 =(10201, -9409, 8649, -7921, 7225, -6561, 5929, -5329, 4761, -4225, 3721, -3249, 2809, -2401, 2025, -1681, 1369, -1089, 841, -625, 441, -289, 169, -81, 25, -1) =5,304

Guest Jul 1, 2020
 #3
avatar+25541 
+2

(b) Compute the sum

\(\Big(a +(2n+1)d\Big)^2- \Big(a + (2n)d\Big)^2 +\Big(a + (2n-1)d\Big)^2 - \Big(a+(2n-2)d\Big)^2 + \cdots + (a+d)^2 - a^2.\).

 

\(\small{ \begin{array}{|rcll|} \hline && \mathbf{\Big(a +(2n+1)d\Big)^2- \Big(a + (2n)d\Big)^2 +\Big(a + (2n-1)d\Big)^2 - \Big(a+(2n-2)d\Big)^2 + \cdots + (a+d)^2 - a^2} \\\\ &=& \Big(a +(2n+1)d\Big)^2 +\Big(a + (2n-1)d\Big)^2+\Big(a + (2n-3)d\Big)^2 + \cdots + \Big(a+(2n-(2n-1))d\Big)^2 \\ && -\Big(a + (2n-0)d\Big)^2- \Big(a+(2n-2)d\Big)^2- \Big(a+(2n-4)d\Big)^2 - \cdots -\Big(a+(2n-(2n))d\Big)^2 \\\\ &=& \mathbf{\sum \limits_{k=1}^{n+1}\Big(~ a+(2k-1)d~\Big)^2 -\sum \limits_{k=1}^{n+1}\Big(~ a+2(k-1)d~\Big)^2} \\ \hline \end{array} }\)

 

\(\begin{array}{|rcll|} \hline && \mathbf{\sum \limits_{k=1}^{n+1}\Big(~ a+(2k-1)d~\Big)^2 -\sum \limits_{k=1}^{n+1}\Big(~ a+2(k-1)d~\Big)^2} \\\\ &=& \sum \limits_{k=1}^{n+1}\Big(~ a+(2k-1)d~\Big)^2 - \Big(~a+2(k-1)d~\Big)^2 \\\\ &=& \sum \limits_{k=1}^{n+1} a^2+2ad(2k-1)+d^2(2k-1)^2-\Big(~ a^2+4ad(k-1)+4d^2(k-1)^2 ~\Big) \\\\ &=& \sum \limits_{k=1}^{n+1} {\color{red}{a^2}}+2ad(2k-1)+d^2(2k-1)^2 {\color{red}{-a^2}}-4ad(k-1)-4d^2(k-1)^2 \\\\ &=& \sum \limits_{k=1}^{n+1} 2ad(2k-1)+d^2(2k-1)^2 -4ad(k-1)-4d^2(k-1)^2 \\\\ &=& \sum \limits_{k=1}^{n+1} {\color{red}{4adk}}-2ad+d^2(2k-1)^2 {\color{red}{-4adk}}+4ad-4d^2(k-1)^2 \\\\ &=& \sum \limits_{k=1}^{n+1} -2ad+d^2(2k-1)^2 +4ad-4d^2(k-1)^2 \\\\ &=& \sum \limits_{k=1}^{n+1} 2ad+d^2(2k-1)^2 -4d^2(k-1)^2 \\\\ &=& \sum \limits_{k=1}^{n+1}2ad +d^2~\sum \limits_{k=1}^{n+1} (2k-1)^2 -4(k-1)^2 \quad | \quad \sum \limits_{k=1}^{n+1}2ad = 2ad(n+1) \\\\ &=& 2ad(n+1) +d^2\sum \limits_{k=1}^{n+1} (2k-1)^2 -4(k-1)^2 \\\\ &=& 2ad(n+1) +d^2\sum \limits_{k=1}^{n+1} 4k^2-4k+1-4(k^2-2k+1) \\\\ &=& 2ad(n+1) +d^2\sum \limits_{k=1}^{n+1} {\color{red}{4k^2}}-4k+1{\color{red}{-4k^2}}+8k-4 \\\\ &=& 2ad(n+1) +d^2\sum \limits_{k=1}^{n+1} 4k-3 \\\\ &=& 2ad(n+1) +d^2\Big(~4\sum \limits_{k=1}^{n+1}k-\sum \limits_{k=1}^{n+1}3~\Big)\quad | \quad \sum \limits_{k=1}^{n+1}3 = 3(n+1) \\\\ &=& 2ad(n+1) +d^2\Big(~4\sum \limits_{k=1}^{n+1}k-3(n+1)~\Big) \\\\ &=& 2ad(n+1) +d^2\Big(~4\sum \limits_{k=1}^{n+1}k-3(n+1)~\Big) \quad | \quad \sum \limits_{k=1}^{n+1}k = \dfrac{1+(n+1)}{2}(n+1) \\\\ &=& 2ad(n+1) +d^2\Big(~4\dfrac{(n+2))}{2}(n+1)-3(n+1)~\Big) \\\\ &=& 2ad(n+1) +d^2\Big(~2(n+2)(n+1)-3(n+1)~\Big) \\\\ &=& 2ad(n+1) +d^2(n+1)\Big(~2(n+2)-3~\Big) \\\\ &=& 2ad(n+1) +d^2(n+1)(2n+4-3) \\\\ &=& 2ad(n+1) +d^2(n+1)(2n+1) \\ &=& \mathbf{d(n+1)\Big(2a+d(2n+1)\Big)} \\ \hline \end{array}\)

 

\(\mathbf{\Big(a +(2n+1)d\Big)^2- \Big(a + (2n)d\Big)^2 +\Big(a + (2n-1)d\Big)^2 - \Big(a+(2n-2)d\Big)^2 + \cdots + (a+d)^2 - a^2} \\ = \mathbf{d(n+1)\Big(2a+d(2n+1)\Big)}\)

 

laugh

 Jul 1, 2020

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