We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website. cookie policy and privacy policy.
 
+0  
 
0
50
4
avatar

What is the sum of this sequence?
1000^2 + 999^2 - 998^2 - 997^2 +996^2 + 995^2 - .........+ 4^2 + 3^2 - 2^2 - 1^2. Any help would be great. Thank you.

 Mar 24, 2019
 #1
avatar+70 
+2

Let's do this a different way! 
Group the terms in pairs. 
1000² - 999² = (1000+999)(1000-999) = 1999. 
998² - 997² = (998 + 997)(998-997) = 1995. 



2²-1² = (2+1)(2-1) = 3. 

So we have to sum the arithmetic progression 
3 + 7 + ... + (4n-1) 
to 500 terms, since 1999 is the 500th term of 
this progression. 
So, how do we do this? 
Note that 
3 + 1999 = 2002 
7 + 1995 = 2002 
11 + 1991 = 2002 



1999 + 3 = 2002 
Call the sum of each column S 
Then S + S = 500*2002 
S = 500*1001 = 500500.

 Mar 24, 2019
 #2
avatar
+2

1000^2 + 999^2 - 998^2 - 997^2..........etc.

Why did you subtract the first 2 terms?

 

Every 4 terms form an arithmetic sequence as follows:
7,988, 7,956, 7,924, 7,892........etc., where:
First term =7,988
Common difference = - 32
Number of terms =1,000 / 4 = 250
SUM = N/2 * [2*F + (N - 1)* D]
         =250/2*[2*7,988 + (250 - 1) * - 32]
         =125 * [15,976 + (249 * - 32)]
         =125 * [15,976 - 7,968]
         =125 * [8,008]
         = 1,001,000

 Mar 24, 2019
 #3
avatar
+1

∑[- 4(8n - 2005), n, 1, 250] = 1,001,000

 Mar 24, 2019
 #4
avatar+21978 
+3

What is the sum of this sequence?
1000^2 + 999^2 - 998^2 - 997^2 +996^2 + 995^2 - .........+ 4^2 + 3^2 - 2^2 - 1^2.

 

\(\begin{array}{|rcll|} \hline && \mathbf{1000^2 + 999^2 - 998^2 - 997^2 +996^2 + 995^2 - \ldots + 4^2 + 3^2 - 2^2 - 1^2} \\ &=& \sum \limits_{k = 1}^{250} \Big( (4k)^2+ (4k-1)^2- (4k-2)^2 -(4k-3)^2 \Big) \\ &=& \sum \limits_{k = 1}^{250} \Big( 16k^2 +16k^2-8k+1-16k^2+16k-4 -16k^2+24k-9 \Big) \\ &=& \sum \limits_{k = 1}^{250} \Big( -8k+1 +16k-4 +24k-9 \Big) \\ &=& \sum \limits_{k = 1}^{250} \Big( 32k-12 \Big) \\ &=& \sum \limits_{k = 1}^{250}32k-\sum \limits_{k = 1}^{250}12 \\ &=& 32\sum \limits_{k = 1}^{250}k- 12\cdot 250 \\ &=& 32\left(\dfrac{1+250}{2}\right)\cdot 250- 12\cdot 250 \\ &=& 16\cdot 251 \cdot 250- 12\cdot 250 \\ &=& 250( 16\cdot 251 - 12) \\ &=& 250\cdot 4004 \\ &\mathbf{=}& \mathbf{1001000} \\ \hline \end{array}\)

 

laugh

 Mar 25, 2019

14 Online Users

avatar
avatar
avatar