Rewrite the following sum with the index of summation

[Slicktacker]

Hi, I don't understand this problem at all:

Rewrite the following sum with the index of summation starting at 3 in summation notation:

\sum_{i=1}^{6}(5+3i)

I know that the sum is 93 but I'm not sure what to do...

Thanks for the help!

 

[arildno]

Let j=i+2.

Since 1<=i<=6, clearly we have 3<=j<=8

Use this information to rewrite the the sum in terms of summation index j

 

[Muzza]

My guess is that the author of the problem wants you to find (possibly integer?) x, y such that

\sum_{i=1}^{6}(5+3i) = \sum_{k = 3}^{6}(x + yk).

But it's not really clear if upper limit of the sum has to be 6 (in that case, it's impossible to find integer solutions for x and y).

 

[Slicktacker]

I tried the following:

\sum_{i=3}^{4}(36+3i)

and it worked. Weird...

 

[HallsofIvy]

As arildno said: let j= i+ 2 (so that i= j- 2). The lower bound on the sum was i= 1 which corresponds to j= 1+2= 3. The upper bound was i= 6 which corresponds to j= 6+ 2= 8. In addition, the "summand" is 5+ 3i= 5+ 3(j- 2)= 5+ 3j- 6= 3j-1. The sum is
/sum_{j=3}^{8}{3j- 1}

 

[dannyboy]

golden rule is, whatever you add to the index of summation, subtract from the variable. so, say you have an infinite series beginning with i=0 for the function i^2, and you are told to express this as a sum starting from i=3, you would subtract 3 from the variable; i.e the sum from i=0 to infinity of i^2 is identical to the sum from i=3 to infinity of (i-3)^2.

DB

 

[JonF]

\sum_{i=1}^{6}(5+3i) = \sum_{i=1+2}^{6+2}(5+3(i-2)) = \sum_{i=3}^{8}(5+3i-6) = \sum_{i=3}^{8}3i-1

 

[maverick280857]

golden rule is, whatever you add to the index of summation, subtract from the variable. so, say you have an infinite series beginning with i=0 for the function i^2, and you are told to express this as a sum starting from i=3, you would subtract 3 from the variable; i.e the sum from i=0 to infinity of i^2 is identical to the sum from i=3 to infinity of (i-3)^2.

DB

This is similar to translating the "origin" (the starting value of the index). This is one of the properties of summations (but in effect it is no big deal since you're simply changing the way you add the quantities, but that does not mean you're changing the sum). This is however a very useful tip at times as dannyboy and JohnF have effectively demonstrated.

A quantity that does not depend on the index of the summation can be taken out of it. This means that \sum_{i = i_{min}}^{i_{max}}ax_{i} equals a\sum_{i = i_{min}}^{i_{max}}x_{i}. This (rather trivial sounding analogy but very useful at times) + dannyboy's tip + some other properties of summations coupled with some ingenuity can really help you in problems :-)

Cheers
Vivek

 

[Alkatran]

Hi, I don't understand this problem at all:

Rewrite the following sum with the index of summation starting at 3 in summation notation:

\sum_{i=1}^{6}(5+3i)

I know that the sum is 93 but I'm not sure what to do...

Thanks for the help!

Couldn't you put the minimum and maximums as the same number, solve the original equation, and place the solution as the series to be summed. (I'll apologize for the lack of correct terms here as I learned my math in french)