Alternatively, we could observe that every even term is 0, so we only need to generate odd terms. We could have expressed the `b_n` term as:

`b_n=-5/(n pi)(cos n pi-1)`

`=-5/(n pi)((-1)^n-1)`

`=10/(n pi)`, if `n` is odd, and `0` if `n` is even.

To generate odd numbers for our series, we need to use:

`b_n=10/((2n-1)pi),\ \ \ n=1,2,3...`

We also need to generate only odd numbers for the sine terms in the series, since the even ones will be 0.

So the required series this time is:

`f(t)=a_0/2` `+sum_(n=1)^oo a_n\ cos {:(n pi t)/L :}` `+sum_(n=1)^oob_n\ sin {:(n pi t)/L:}`

`=5/2` `+sum_(n=1)^oo(0)\ cos {:(n pi t)/4:}` `+sum_(n=1)^oo 10/((2n-1)pi) sin {:((2n-1)pi t)/4:}`

`=2.5` `+10/pi sum_(n=1)^oo 1/((2n-1)) sin {:((2n-1)pi t)/4:}`

The first four terms series are once again:

`f(t)=2.5+10/pi(sin {:(pi t)/4:}` `+1/3 sin {:(3 pit)/4:}` `+1/5 sin {:(5pi t)/4:}` `{:+1/7 sin {:(7pi t)/4:}...)`