If f(x) be a differentiable function satisfying f(x)+f(x+12)=1 ∀x∈ – InterviewSolution

1.	If f(x) be a differentiable function satisfying f(x)+f(x+12)=1 ∀x∈R and g(x)=x∫0f(t)dt. Then the value of ∞∑n=2⎛⎜⎜⎜⎜⎜⎝8n∑k=1(g(x+k2)−g(x+k)⎞⎟⎟⎟⎟⎟⎠ is
Answer» If $f (x)$ be a differentiable function satisfying $f (x) + f (x + \frac{1}{2}) = 1 \forall x \in R$ and $g (x) = x \int 0 f (t) d t$ . Then the value of $\infty \sum n = 2 ⎛ ⎜⎜⎜⎜⎜ ⎝ \frac{8}{n \sum k = 1 (g (x + k^{2}) - g (x + k)} ⎞ ⎟⎟⎟⎟⎟ ⎠$ is

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