# Thread: help finding sum of series

1. ## 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...

got 2 series.

1.

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

(ok those are just squares)

need to find sum...
$\displaystyle 1^2-2^2+3^2-4^2+\cdots+99^2-100^2$

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

finding the difference in pairs now giving 50 terms.

$\displaystyle -3-7-11-\cdots$

Use sum of an arithmetic sequence

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

where n = number of terms = 50

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

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

Are you familiar with these summation formulas?

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

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

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

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

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

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

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

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

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

finding the difference in pairs now giving 50 terms.

$\displaystyle -3-7-11-\cdots$

Use sum of an arithmetic sequence

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

where n = number of terms = 50

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

$\displaystyle S_{50} = \frac{50}{2}(2(-3)+(50-1)(-4)) = \cdots$
thats not coming out to be correct...its -5050.
refer to soroban's solution.

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

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

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

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

$\displaystyle S_{50} = 25(-202)$

$\displaystyle S_{50} = -5050$

Worked for me!

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

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

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

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

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

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

$\displaystyle S_{50} = 25(-202)$

$\displaystyle S_{50} = -5050$

Worked for me!
hey! so sorry. i think i went wrong with the calc.(-calculation-)

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

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

so here $\displaystyle n = 50$ so the answer $\displaystyle S = 51!-1$.
hey! can we similarily prove the sum $\displaystyle n^{2}.n!$???