Using these identities:
sinh(mx+x)=cosh(mx)sinh(x)+sinh(mx)cosh(x)
cosh(mx+x)=cosh(mx)cosh(x)+sinh(mx)sinh(x) Express the following sums in terms of just cosh((n+1)x), sinh((n+1)x), cosh(x) and sinh(x):
Cn=cosh(1x)+cosh(2x)+cosh(3x)+...cosh(nx)
- Sn=sinh(1x)+sinh(2x)+sinh(3x)+...sinh(nx)
Looking for full worked answer.
For the first one, consider that $$\sinh(n+1)x-\sinh(n-1)x=2\cosh nx\sinh x$$ So if you now consider $$2C_n\sinh x=(\sinh 2x-\sinh 0)+(\sinh 3x-\sinh x)+(\sinh 4x-\sinh 2x)+...$$ This is a telescoping sum which leaves you with $$2C_n\sinh x=\sinh (n+1)x+\sinh nx-\sinh x$$ and hence you have the required sum.
The other one can be done in a similar way.