Let $$f(x)$$ be a function such that $$f(x$$−$$1)+$$ $$f(x+1)=2f(x)$$ for all x∈R . If $$f(3)=5$$ then ∑$$r=010$$ $$f(3+8r)$$ is equal to

Question

Let $$f(x)$$ be a function such that $$f(x$$−$$1)+$$ $$f(x+1)=2f(x)$$ for all x∈R . If $$f(3)=5$$ then ∑$$r=010$$ $$f(3+8r)$$ is equal to

Infinity Learn · Accepted Answer

The given equation is $$f(x$$ – $$1)$$ $$+$$ $$f(x$$ $$+$$ $$1)$$ $$=$$ $$2$$ f (x)          …(i)Replace$$ x by x –$$1$$ in  $$(i) $$f(x$$ – $$2)$$ $$+$$ $$f(x)$$ $$=$$ $$2$$ $$f(x$$ $$-$$ $$1)$$         …(ii)Replace$$ x by x $$+1$$ in $$(ii) $$f(x)$$ $$+$$ $$f(x$$ $$+$$ $$2)$$ $$=$$ $$2$$ $$f(x+1)$$           …(iii)Adding$$ $$(ii) and (iii), we have $$f(x$$ – $$2)$$ $$+$$ $$f(x$$ $$+$$ $$2)$$ $$+$$ $$2$$ $$f(x)$$ $$=$$ $$2$$ $$($$ $$f(x$$ – $$1)$$ $$+$$ $$f(x$$ $$+$$ $$1))f(x$$ – $$2)$$ $$+$$ $$f(x$$ $$+$$ $$2)$$ $$+$$ $$2$$ $$f(x)$$ $$=$$ $$2$$ $$($$ $$f(x$$ – $$1)$$ $$+$$ $$f(x$$ $$+$$ $$1))$$   $$=$$ $$2$$ $$2$$ $$f(x)$$ $$=$$ $$2f(x)$$⇒$$f(x$$ –$$2)$$ $$=$$– f (x$$ $$+$$ $$2) Replacing x by x $$+$$ $$2$$, we $$havef(x)$$ $$=$$ –$$f(x$$ $$+$$ $$4)$$ $$=$$ – $$($$ f (x$$ $$+$$ $$8)) $$=$$ $$f(x$$ $$+$$ $$8)$$∑$$r=010$$ $$f(3+8r)=11$$×$$f(3)=11$$×$$5=55$$

Let f(x) be a function such that $f (x - 1) +$ $f (x + 1) = \sqrt{2} f (x)$ for all $x \in R$ . If $f (3) = 5 then \sum_{r = 0}^{10} f (3 + 8 r)$ is equal to

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Let f(x) be a function such that f(x−1)+ f(x+1)=2f(x) for all x∈R . If f(3)=5 then ∑r=010 f(3+8r) is equal to

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Let f(x) be a function such that $f (x - 1) +$ $f (x + 1) = \sqrt{2} f (x)$ for all $x \in R$ . If $f (3) = 5 then \sum_{r = 0}^{10} f (3 + 8 r)$ is equal to