Let `f` be an invertible real function. Write `(f^(-1)\ of)(1)+(f^(-1)\ of)(2)++(f^(-1)\ of)(100)dot`

Explanation

`{f}^{-1} circ {f}={I}`, where `I` is an identity function.


So,



`({f}^{-1} circ {f})(1)+({f}^{-1} circ {f})(2)+ldots+({f}^{-1} circ {f})(100) `


`Rightarrow {I}(1)+{I}(2)+ldots .+{I}(100) `


`Rightarrow 1+2+ldots+100 quad[{As} {I}_{({x})}={x}, forall_{{x}} in {R}]`



Sum of first `n` natural numbers `=frac{n(n+1)}{2}`


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