+0

# Help plzz!!!

0
48
7

Compute the sum:

101^2-97^2+93^2-89^2+.....+5^2-1^2

Thank you

May 2, 2020

#5
+24909
+2

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

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

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

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

May 4, 2020
edited by heureka  May 4, 2020

#1
+29978
+2

Heureka has also answered this previously I think.

May 2, 2020
#2
+111360
+1

Let's see if we can find a pattern

101^2  - 97^2  =  792

93^2  - 89^2  =  728

85^2  - 81^2  = 664

....

21^2 - 17^2  = 152

13^2 - 9^2  =  88

5^2 - 1^2  =  24

And note that

(792 + 24)  =  ( 728 + 88) =  ( 664 + 152)  =   816

5    13      21      29      37     45      53     61     69    77    85    93    101

1     9       17     25       33     41      49     57     65    73    81    89      97

So  we  will have 6 pairs  of  equal sums of 816   plus  the sum of  (53^2 - 49^2)  = 408

So....the sum is     6 (816) + 408  =  5304

May 2, 2020
edited by CPhill  May 2, 2020
edited by CPhill  May 2, 2020
#4
0

Thanks a lot CPhill

Guest May 2, 2020
#3
+1

sum_(n=1)^26 (-1)^(n + 1) (105 - 4 n)^2 = 5304

May 2, 2020
#5
+24909
+2

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

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

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

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

heureka May 4, 2020
edited by heureka  May 4, 2020
#6
+111360
+1

Thanks, heureka.....!!!

I didn't see that we could express this  as a difference of summations....I like this method  !!!!

CPhill  May 4, 2020
#7
+24909
+2

Thank you, CPhill !

heureka  May 5, 2020