Summation rules and properties

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I am trying to find the sum of this -

$$sum_r=1^100 ( 1 + 2r + 0.3^r ) $$

I know roughly how I am supposed to do. First I distribute the summation across the 3 values.

Then I got stuck $2r$ and $0.3^r$

Both are similar if I understand, so I will only ask for one of them ,

$sum_r=1^100 (2r)$

how am I suppose to use the property to solve this just like $sum_r=1^100 = 100(1) = 100 $

Am I right to say -

$sum_r=1^100 (2r) = 2(1)(100) $ ? But I doubt myself as the r value changes . So I think this is wrong.

summation

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edited Jul 22 at 17:20

asked Jul 22 at 17:18

user185692

1045

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Try separating the sum into sums of $1$, $2r$, and $0.3^r$ then solving each independently.
  – Shrey Joshi
  Jul 22 at 17:20
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $sum_r=1^100 (2r) = (1+2+cdots+99+100) + (100+99+cdots+2+1) $ $= (1+100)+(2+99)+cdots+(99+2)+(100+1) = 100 times 101$
  – Henry
  Jul 22 at 17:25
  
  

  
  
  
  

add a comment | 

up vote
0
down vote

favorite

I am trying to find the sum of this -

$$sum_r=1^100 ( 1 + 2r + 0.3^r ) $$

I know roughly how I am supposed to do. First I distribute the summation across the 3 values.

Then I got stuck $2r$ and $0.3^r$

Both are similar if I understand, so I will only ask for one of them ,

$sum_r=1^100 (2r)$

how am I suppose to use the property to solve this just like $sum_r=1^100 = 100(1) = 100 $

Am I right to say -

$sum_r=1^100 (2r) = 2(1)(100) $ ? But I doubt myself as the r value changes . So I think this is wrong.

summation

share|cite|improve this question

edited Jul 22 at 17:20

asked Jul 22 at 17:18

user185692

1045

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Try separating the sum into sums of $1$, $2r$, and $0.3^r$ then solving each independently.
  – Shrey Joshi
  Jul 22 at 17:20
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $sum_r=1^100 (2r) = (1+2+cdots+99+100) + (100+99+cdots+2+1) $ $= (1+100)+(2+99)+cdots+(99+2)+(100+1) = 100 times 101$
  – Henry
  Jul 22 at 17:25
  
  

  
  
  
  

add a comment | 

up vote
0
down vote

favorite

up vote
0
down vote

favorite

I am trying to find the sum of this -

$$sum_r=1^100 ( 1 + 2r + 0.3^r ) $$

I know roughly how I am supposed to do. First I distribute the summation across the 3 values.

Then I got stuck $2r$ and $0.3^r$

Both are similar if I understand, so I will only ask for one of them ,

$sum_r=1^100 (2r)$

how am I suppose to use the property to solve this just like $sum_r=1^100 = 100(1) = 100 $

Am I right to say -

$sum_r=1^100 (2r) = 2(1)(100) $ ? But I doubt myself as the r value changes . So I think this is wrong.

summation

share|cite|improve this question

edited Jul 22 at 17:20

asked Jul 22 at 17:18

user185692

1045

I am trying to find the sum of this -

$$sum_r=1^100 ( 1 + 2r + 0.3^r ) $$

I know roughly how I am supposed to do. First I distribute the summation across the 3 values.

Then I got stuck $2r$ and $0.3^r$

Both are similar if I understand, so I will only ask for one of them ,

$sum_r=1^100 (2r)$

how am I suppose to use the property to solve this just like $sum_r=1^100 = 100(1) = 100 $

Am I right to say -

$sum_r=1^100 (2r) = 2(1)(100) $ ? But I doubt myself as the r value changes . So I think this is wrong.

summation

share|cite|improve this question

edited Jul 22 at 17:20

asked Jul 22 at 17:18

user185692

1045

share|cite|improve this question

share|cite|improve this question

edited Jul 22 at 17:20

edited Jul 22 at 17:20

edited Jul 22 at 17:20

asked Jul 22 at 17:18

user185692

1045

asked Jul 22 at 17:18

user185692

1045

asked Jul 22 at 17:18

user185692

1045

1045

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Try separating the sum into sums of $1$, $2r$, and $0.3^r$ then solving each independently.
  – Shrey Joshi
  Jul 22 at 17:20
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $sum_r=1^100 (2r) = (1+2+cdots+99+100) + (100+99+cdots+2+1) $ $= (1+100)+(2+99)+cdots+(99+2)+(100+1) = 100 times 101$
  – Henry
  Jul 22 at 17:25
  
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Try separating the sum into sums of $1$, $2r$, and $0.3^r$ then solving each independently.
  – Shrey Joshi
  Jul 22 at 17:20
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $sum_r=1^100 (2r) = (1+2+cdots+99+100) + (100+99+cdots+2+1) $ $= (1+100)+(2+99)+cdots+(99+2)+(100+1) = 100 times 101$
  – Henry
  Jul 22 at 17:25
  
  

  
  
  
  

Try separating the sum into sums of $1$, $2r$, and $0.3^r$ then solving each independently.
– Shrey Joshi
Jul 22 at 17:20

Try separating the sum into sums of $1$, $2r$, and $0.3^r$ then solving each independently.
– Shrey Joshi
Jul 22 at 17:20

$sum_r=1^100 (2r) = (1+2+cdots+99+100) + (100+99+cdots+2+1) $ $= (1+100)+(2+99)+cdots+(99+2)+(100+1) = 100 times 101$
– Henry
Jul 22 at 17:25

$sum_r=1^100 (2r) = (1+2+cdots+99+100) + (100+99+cdots+2+1) $ $= (1+100)+(2+99)+cdots+(99+2)+(100+1) = 100 times 101$
– Henry
Jul 22 at 17:25

add a comment | 

2 Answers
2

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3
down vote



accepted

You should try breaking up the summation into parts so that you can evaluate them easily

$displaystyle sum_r=1^100 ( 1 + 2r + 0.3^r )$

$=displaystyle sum_r=1^1001+sum_r=1^1002r+sum_r=1^100(0.3)^r$

$displaystyle = 100+2sum_r=1^100r+sum_r=1^100(0.3)^r$

$displaystyle= 100+2cdotfrac100(101)2+ frac1-(0.3)^1001-0.3$

$=displaystyle 100+10100+1.428 $

$=10201.4285$

EDIT:

You should look up geometric series to understand the summation of $(0,3)^r $

share|cite|improve this answer

answered Jul 22 at 17:24

The Integrator

3,0262725

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up vote
2
down vote

Hint:
$$sum_r=1^1001+2r+0.3^r=left(sum_r=1^1001right)+left(sum_r=1^1002rright)+left(sum_r=1^1000.3^rright)$$

Solution:

$$sum_r=1^1001+2r+0.3^r=left(sum_r=1^1001right)+left(sum_r=1^1002rright)+left(sum_r=1^1000.3^rright)$$
$$left(sum_r=1^1001right)=100$$
$$left(sum_r=1^1002rright)=2sum_r=1^100r=100(101)=10100$$
Since $1+2+3+...+r=r(r+1)/2$
$$left(sum_r=1^1000.3^rright)=frac1-0.3^1001-0.3$$
The sum is about $10201.428$

share|cite|improve this answer

answered Jul 22 at 17:29

Shrey Joshi

1389

add a comment | 

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2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
3
down vote



accepted

You should try breaking up the summation into parts so that you can evaluate them easily

$displaystyle sum_r=1^100 ( 1 + 2r + 0.3^r )$

$=displaystyle sum_r=1^1001+sum_r=1^1002r+sum_r=1^100(0.3)^r$

$displaystyle = 100+2sum_r=1^100r+sum_r=1^100(0.3)^r$

$displaystyle= 100+2cdotfrac100(101)2+ frac1-(0.3)^1001-0.3$

$=displaystyle 100+10100+1.428 $

$=10201.4285$

EDIT:

You should look up geometric series to understand the summation of $(0,3)^r $

share|cite|improve this answer

answered Jul 22 at 17:24

The Integrator

3,0262725

add a comment | 

up vote
3
down vote



accepted

You should try breaking up the summation into parts so that you can evaluate them easily

$displaystyle sum_r=1^100 ( 1 + 2r + 0.3^r )$

$=displaystyle sum_r=1^1001+sum_r=1^1002r+sum_r=1^100(0.3)^r$

$displaystyle = 100+2sum_r=1^100r+sum_r=1^100(0.3)^r$

$displaystyle= 100+2cdotfrac100(101)2+ frac1-(0.3)^1001-0.3$

$=displaystyle 100+10100+1.428 $

$=10201.4285$

EDIT:

You should look up geometric series to understand the summation of $(0,3)^r $

share|cite|improve this answer

answered Jul 22 at 17:24

The Integrator

3,0262725

add a comment | 

up vote
3
down vote



accepted

up vote
3
down vote



accepted

You should try breaking up the summation into parts so that you can evaluate them easily

$displaystyle sum_r=1^100 ( 1 + 2r + 0.3^r )$

$=displaystyle sum_r=1^1001+sum_r=1^1002r+sum_r=1^100(0.3)^r$

$displaystyle = 100+2sum_r=1^100r+sum_r=1^100(0.3)^r$

$displaystyle= 100+2cdotfrac100(101)2+ frac1-(0.3)^1001-0.3$

$=displaystyle 100+10100+1.428 $

$=10201.4285$

EDIT:

You should look up geometric series to understand the summation of $(0,3)^r $

share|cite|improve this answer

answered Jul 22 at 17:24

The Integrator

3,0262725

You should try breaking up the summation into parts so that you can evaluate them easily

$displaystyle sum_r=1^100 ( 1 + 2r + 0.3^r )$

$=displaystyle sum_r=1^1001+sum_r=1^1002r+sum_r=1^100(0.3)^r$

$displaystyle = 100+2sum_r=1^100r+sum_r=1^100(0.3)^r$

$displaystyle= 100+2cdotfrac100(101)2+ frac1-(0.3)^1001-0.3$

$=displaystyle 100+10100+1.428 $

$=10201.4285$

EDIT:

You should look up geometric series to understand the summation of $(0,3)^r $

share|cite|improve this answer

answered Jul 22 at 17:24

The Integrator

3,0262725

share|cite|improve this answer

share|cite|improve this answer

answered Jul 22 at 17:24

The Integrator

3,0262725

answered Jul 22 at 17:24

The Integrator

3,0262725

answered Jul 22 at 17:24

The Integrator

3,0262725

3,0262725

add a comment | 

add a comment | 

up vote
2
down vote

Hint:
$$sum_r=1^1001+2r+0.3^r=left(sum_r=1^1001right)+left(sum_r=1^1002rright)+left(sum_r=1^1000.3^rright)$$

Solution:

$$sum_r=1^1001+2r+0.3^r=left(sum_r=1^1001right)+left(sum_r=1^1002rright)+left(sum_r=1^1000.3^rright)$$
$$left(sum_r=1^1001right)=100$$
$$left(sum_r=1^1002rright)=2sum_r=1^100r=100(101)=10100$$
Since $1+2+3+...+r=r(r+1)/2$
$$left(sum_r=1^1000.3^rright)=frac1-0.3^1001-0.3$$
The sum is about $10201.428$

share|cite|improve this answer

answered Jul 22 at 17:29

Shrey Joshi

1389

add a comment | 

up vote
2
down vote

Hint:
$$sum_r=1^1001+2r+0.3^r=left(sum_r=1^1001right)+left(sum_r=1^1002rright)+left(sum_r=1^1000.3^rright)$$

Solution:

$$sum_r=1^1001+2r+0.3^r=left(sum_r=1^1001right)+left(sum_r=1^1002rright)+left(sum_r=1^1000.3^rright)$$
$$left(sum_r=1^1001right)=100$$
$$left(sum_r=1^1002rright)=2sum_r=1^100r=100(101)=10100$$
Since $1+2+3+...+r=r(r+1)/2$
$$left(sum_r=1^1000.3^rright)=frac1-0.3^1001-0.3$$
The sum is about $10201.428$

share|cite|improve this answer

answered Jul 22 at 17:29

Shrey Joshi

1389

add a comment | 

up vote
2
down vote

up vote
2
down vote

Hint:
$$sum_r=1^1001+2r+0.3^r=left(sum_r=1^1001right)+left(sum_r=1^1002rright)+left(sum_r=1^1000.3^rright)$$

Solution:

$$sum_r=1^1001+2r+0.3^r=left(sum_r=1^1001right)+left(sum_r=1^1002rright)+left(sum_r=1^1000.3^rright)$$
$$left(sum_r=1^1001right)=100$$
$$left(sum_r=1^1002rright)=2sum_r=1^100r=100(101)=10100$$
Since $1+2+3+...+r=r(r+1)/2$
$$left(sum_r=1^1000.3^rright)=frac1-0.3^1001-0.3$$
The sum is about $10201.428$

share|cite|improve this answer

answered Jul 22 at 17:29

Shrey Joshi

1389

Hint:
$$sum_r=1^1001+2r+0.3^r=left(sum_r=1^1001right)+left(sum_r=1^1002rright)+left(sum_r=1^1000.3^rright)$$

Solution:

$$sum_r=1^1001+2r+0.3^r=left(sum_r=1^1001right)+left(sum_r=1^1002rright)+left(sum_r=1^1000.3^rright)$$
$$left(sum_r=1^1001right)=100$$
$$left(sum_r=1^1002rright)=2sum_r=1^100r=100(101)=10100$$
Since $1+2+3+...+r=r(r+1)/2$
$$left(sum_r=1^1000.3^rright)=frac1-0.3^1001-0.3$$
The sum is about $10201.428$

share|cite|improve this answer

answered Jul 22 at 17:29

Shrey Joshi

1389

share|cite|improve this answer

share|cite|improve this answer

answered Jul 22 at 17:29

Shrey Joshi

1389

answered Jul 22 at 17:29

Shrey Joshi

1389

answered Jul 22 at 17:29

Shrey Joshi

1389

1389

add a comment | 

add a comment | 

 

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