Sigma Notation: sum[i=1 to 6] [n cos(n pi)]

[d1zz]

I am currently stuck on this question. Please help.

. . . . . . .6
. . .(Sigma Notation) n cos(n pi)
. . . . . .i=1

Thanks!

 

[tkhunny]

There are only six terms. What's stopping you from listing them?

1*cos(pi) + 2*cos(2pi) + 3*cos(3pi) + etc.

Then evaluate the cosines...

1*(-1) + 2*(1) + 3*(-1) + etc.

Simplify

-1 + 2 - 3 + 4 - 5 + 6

Simplify some more...

 

[stapel]

What is the actual question? Are you supposed to expand this? List the last two terms? Evaluate the whole thing? Or something else?

When you reply, please show everything you have tried so far. Thank you.

Eliz.

 

[d1zz]

Sorry

They way I see the question is:

Since Cos pi= -1

and n having six terms from 1 to 6.

1(-1(1)) + 2((-1(2)) + 3(-1(3)....

That would give me a totally different answer from the second post...

 

[stapel]

Since Cos pi= -1....

You gave the summand as "n cos(n pi)", but you are now evaluating it as "n cos(pi) n = n² cos(pi)". (Otherwise, your second term, to use one example, would have been "2 cos(2pi) = 2(1) = 2", as the second post listed.)

Was your initial post incorrect...?

Eliz.

 

[tkhunny]

Yes, please, let's make up our minds what the problem statement is.