Let $f:\mathbb{R}\rightarrow\mathbb{R}$ and $g:\mathbb{R}\rightarrow\mathbb{R}$ be differentiable functions such that $g(f(x))=x$ and $f(x)=2x+\cos x+\sin^2 x$. Then $\displaystyle \sum^{99}_{n=1}g\bigg(1+(2n-1)\pi\bigg)$ is equal to …?
Solution I try:
$$g(1+\pi)+g(1+3\pi)+\cdots \cdots +g(1+97\pi).$$
If $f(x)$ and $g(x)$ are inverses of each other, then
$$f(g(x))=g(f(x))=x.$$
From $g(f(x))=x\implies g(2x+\cos x+\sin^2 x)= x$.
How do I find series sum, help me.
Hint: $$g(x)=f^{-1}(x)$$ and $$f(\frac{n\pi}2)=1+n\pi$$ thus $$g(1+n\pi)=f^{-1}(1+n\pi)=\frac{n\pi}2$$ for all odd $n$.
By the way, $$\sum_{n=1}^kn=\frac{k(k+1)}2$$