Sigma notation problem

[Imum Coeli]

Hi

I have a problem with sigma notation. I thought I knew how to use it, but I have come across questions I can't solve. I'm not sure if this is the right place but any help would be appreciated. Also I don't know to use the symbolic notation so I'll have to try and use words instead of the sigma notation itself.

Question:

Suppose the sum of a_ifor i=1,...,10 is 15 and the sum of a_i for i=1,...,20 is -3. Evaluate the following sums.

a)
The sum of (a_j-2 + a_j+8) for j = 3,...,12

and

b)
The sum of (The sum of (1+a_5i+j) for j=1,...,5) for i=0,1

Notes:

I can see that it is a question of changing the indices but I don't know how to make it work.
Take a) for example. As I think I want to change j from 3,...,12 to 1,...,10 I end up with

The sum of a_j for j=1,...,10 plus The sum of a_j+10 for j=1,...,10

The first term is good but I don't know what to do with the j+10 subscript in the second term.

I know the notation is bad and hard to read but I'd be very grateful for any pointers. Thanks.

 

[Imum Coeli]

You've lost me.
Does that mean sum for i=11,...,20 is -15 - 3 = -18 ?

I don't even know. I think it is a weird question. But calling this gibberish (the sum of a_ifor i=1,...,10 is 15), A, and calling this (sum of a_i for i=1,...,20 is -3), B, then A+B=12.

 

[pka]

Question:
Suppose the sum of a_ifor i=1,...,10 is 15 and the sum of a_i for i=1,...,20 is -3. Evaluate the following sums.
a)
The sum of (a_j-2 + a_j+8) for j = 3,...,12

\(\displaystyle \sum\limits_{j = 3}^{12} {\left( {{a_{j - 2}} + {a_{j + 8}}} \right)} = \sum\limits_{k = 1}^{10} {{a_k}} + \sum\limits_{n = 11}^{20} {{a_n}}=~?\)

 

[soroban]

Hello, Imum Coeli!

Here's the first one.

\(\displaystyle \displaystyle\text{Given:}\:\sum^{10}_{i=1} a_1 \,=\,15 \qquad \sum^{20}_{i=1}a_i \,=\,-3\)

\(\displaystyle \displaystyle\text{(a) Evaluate: }\:\sum^{12}_{j=3}\left(a_{j-2} + a_{j+8}\right)\)

\(\displaystyle \displaystyle\sum^{12}_{j=3}\left(a_{_{j-2}} + a_{_{j+8}}\right)\)
. . . . \(\displaystyle \displaystyle=\;\sum^{10}_{k=1}(a_{_k} + a _{_{k+10}}) \)

. . . . \(\displaystyle \displaystyle =\;\sum^{10}_{k=1}a_{{k}} + \sum^{10}_{k=1}a_{_{k+10}} \)

. . . . \(\displaystyle \displaystyle =\;\left(a_{_1} + a_{_2} + a_{_3} +\cdots + a_{_{10}}\right) + \left(a_{_{11}} + a_{_{12}} + a_{+{13}} + \cdots + a_{_{20}}\right) \)

. . . . \(\displaystyle \displaystyle =\;\sum^{20}_{i=1}a_i\)

. . . . \(\displaystyle =\;-3\)

 

[Imum Coeli]

Wow that is really easy. Thanks so much. By this reasoning b) must be 25. Thanks again.