Question 1083331
<pre>
Let S = the sum of the first n terms of the given sequence.

The nth term will have coefficient (2n-1) and (n-1)st power of x.
So we have:

{{{matrix(1,15,
        S, ""="", 1 ,  ""+"", 3x, ""+"", 5x^2, ""+"", 7x^3, ""+"", ""*""*""*"", ""+"",(2n-1)x^(n-1),    "",        "")}}}

Multiply the sequence through by -x and add term by term to
the given sequence:

{{{matrix(4,15,
        S, ""="", 1 ,  ""+"", 3x, ""+"", 5x^2, ""+"", 7x^3, ""+"", ""*""*""*"", ""+"",(2n-1)x^(n-1),    "",        "",
      -xS, ""="", "",  ""-"",  x, ""-"", 3x^2, ""-"", 5x^3, ""-"", ""*""*""*"", ""-"",(2n-3)x^(n-1), ""-"", (2n-1)x^n,
"Adding,",    "", "",     "", "",    "",   "",    "",   "",    "",          "",    "",           "",    "",        "",
     S-xS, ""="",  1,    "+", 2x, ""+"", 2x^2, ""+"", 2x^3, ""+"", ""*""*""*"", ""+"",     2x^(n-1), ""-"", (2n-1)x^n)}}}

Can you finish from here?:

{{{matrix(1,15,
       
     S(1-x), ""="",  1,    "+", 2x, ""+"", 2x^2, ""+"", 2x^3, ""+"", ""*""*""*"", ""+"",     2x^(n-1), ""-"", (2n-1)x^n)}}}

The n-1 terms between the first and last ones form a geometric 
series.  Use a formula for them, then you can solve for S.

If you have trouble, tell me in the thank-you note form below 
and I'll get back to you by email.  No charge as I do this
for fun.

Edwin</pre>