# help finding sum of series

Printable View

• April 30th 2009, 01:11 AM
adhyeta
help finding sum of series
got 2 series.

1.

12 - 22 + 32 - 42 + ................. 992 - 1002

(ok those are just squares)

2.

1.1! + 2.2! + ....... 50.50!

need to find sum...
• April 30th 2009, 03:07 AM
pickslides
Quote:

Originally Posted by adhyeta
got 2 series.

1.

12 - 22 + 32 - 42 + ................. 992 - 1002

(ok those are just squares)

need to find sum...

$1^2-2^2+3^2-4^2+\cdots+99^2-100^2$

$1-4+9-16+25-36+\cdots+99^2-100^2$

finding the difference in pairs now giving 50 terms.

$-3-7-11-\cdots$

Use sum of an arithmetic sequence

$S_n = \frac{n}{2}(2a+(n-1)d)$

where n = number of terms = 50

a = the first term = -3 and d = the common difference between terms = -4

$S_{50} = \frac{50}{2}(2(-3)+(50-1)(-4)) = \cdots$
• April 30th 2009, 04:22 AM
Soroban
Hello, adhyeta!

Are you familiar with these summation formulas?

. . $\sum^n_{k=1} 1 \;\;= \;n$

. . $\sum^n_{k=1}k \;\;=\;\frac{n(n+1)}{2}$

Quote:

$1)\;\;S \;=\;1^2 - 2^2 + 3^2 - 4^2 + \hdots + 99^2 - 100^2$

$\text{We have: }\;S \;=\;\underbrace{(1^2 + 3^2 + 5^2 + \hdots + 99^2)}_{\text{first 50 odd squares}} \;-\; \underbrace{(2^2+4^2+6^2 + \hdots + 100^2)}_{\text{first 50 even squares}}$

. . $\text{Then: }S \;=\qquad\qquad \sum^{50}_{k=1}(2k-1)^2 \qquad - \qquad\qquad \sum^{50}_{k=1}(2k)^2$

Hence: . $S \;=\;\sum^{50}_{k=1}\bigg[(2k-1)^2 - (2k)^2\bigg] \;=\;\sum^{50}_{k=1}(1-4k)$

. . $= \;\;\sum^{50}_{k=1}\!1 \;-\; 4\!\sum^{50}_{k=1}k \;\;=\;\;50 - 4\,\frac{50\cdot51}{2} \;\;=\;\;\boxed{-5,\!050}$

• April 30th 2009, 06:10 AM
adhyeta
Quote:

Originally Posted by pickslides
$1^2-2^2+3^2-4^2+\cdots+99^2-100^2$

$1-4+9-16+25-36+\cdots+99^2-100^2$

finding the difference in pairs now giving 50 terms.

$-3-7-11-\cdots$

Use sum of an arithmetic sequence

$S_n = \frac{n}{2}(2a+(n-1)d)$

where n = number of terms = 50

a = the first term = -3 and d = the common difference between terms = -4

$S_{50} = \frac{50}{2}(2(-3)+(50-1)(-4)) = \cdots$

thats not coming out to be correct...its -5050.
refer to soroban's solution.
• April 30th 2009, 02:11 PM
pickslides
Quote:

Originally Posted by adhyeta
thats not coming out to be correct...its -5050.
refer to soroban's solution.

$
S_{50} = \frac{50}{2}(2(-3)+(50-1)(-4))$

$

S_{50} = 25(-6+(49)(-4))$

$

S_{50} = 25(-6+-196)$

$

S_{50} = 25(-202)$

$

S_{50} = -5050$

Worked for me!
• April 30th 2009, 03:04 PM
Jester
Quote:

Originally Posted by adhyeta
got 2 series.

1.

12 - 22 + 32 - 42 + ................. 992 - 1002

(ok those are just squares)

2.

1.1! + 2.2! + ....... 50.50!

need to find sum...

Here's the second one. Let

$S_n = 1 \cdot 1! + 2 \cdot 2! + 3 \cdot 3! + \cdots + (n-1) \cdot (n-1)! + n \cdot n!$
$= (2-1) 1! + (3-1)2! + (4-1)3! + \cdots + (n -1) \cdot (n-1)! + (n+1 -1) \cdot n!
$

$= (2! - 1!) + (3! - 2!) + (4! - 3!) + \cdots + (n! - (n-1)!) + ((n+1)! - n!)$
$
= (n+1)! - 1
$

so here $n = 50$ so the answer $S = 51!-1$.
• April 30th 2009, 05:16 PM
adhyeta
Quote:

Originally Posted by pickslides
$
S_{50} = \frac{50}{2}(2(-3)+(50-1)(-4))$

$

S_{50} = 25(-6+(49)(-4))$

$

S_{50} = 25(-6+-196)$

$

S_{50} = 25(-202)$

$

S_{50} = -5050$

Worked for me!

hey! so sorry. i think i went wrong with the calc.(-calculation-)
(Happy)
• May 3rd 2009, 07:16 AM
adhyeta
Quote:

Originally Posted by danny arrigo
Here's the second one. Let

$S_n = 1 \cdot 1! + 2 \cdot 2! + 3 \cdot 3! + \cdots + (n-1) \cdot (n-1)! + n \cdot n!$
$= (2-1) 1! + (3-1)2! + (4-1)3! + \cdots + (n -1) \cdot (n-1)! + (n+1 -1) \cdot n!
$

$= (2! - 1!) + (3! - 2!) + (4! - 3!) + \cdots + (n! - (n-1)!) + ((n+1)! - n!)$
$
= (n+1)! - 1
$

so here $n = 50$ so the answer $S = 51!-1$.

hey! can we similarily prove the sum $n^{2}.n!$???