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Question Number 26281 by kkr1729 last updated on 23/Dec/17

If f(x) is a function satisfying   f(x+y)=f(x) f(y) for all x, y ∈ N  such  that f(1)=3 and Σ_(x=1) ^n  f(x)=120. Then  the value of n is

$$\mathrm{If}\:{f}\left({x}\right)\:\mathrm{is}\:\mathrm{a}\:\mathrm{function}\:\mathrm{satisfying} \\ $$$$\:{f}\left({x}+{y}\right)={f}\left({x}\right)\:{f}\left({y}\right)\:\mathrm{for}\:\mathrm{all}\:{x},\:{y}\:\in\:{N}\:\:\mathrm{such} \\ $$$$\mathrm{that}\:{f}\left(\mathrm{1}\right)=\mathrm{3}\:\mathrm{and}\:\underset{{x}=\mathrm{1}} {\overset{{n}} {\sum}}\:{f}\left({x}\right)=\mathrm{120}.\:\mathrm{Then} \\ $$$$\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:{n}\:\mathrm{is} \\ $$

Answered by mrW1 last updated on 23/Dec/17

f(n+1)=f(1)×f(n)=3f(n)  ⇒G.P. with q=3  3×((3^n −1)/(3−1))=120  3^n =1+80=81=3^4   ⇒n=4

$${f}\left({n}+\mathrm{1}\right)={f}\left(\mathrm{1}\right)×{f}\left({n}\right)=\mathrm{3}{f}\left({n}\right) \\ $$$$\Rightarrow{G}.{P}.\:{with}\:{q}=\mathrm{3} \\ $$$$\mathrm{3}×\frac{\mathrm{3}^{{n}} −\mathrm{1}}{\mathrm{3}−\mathrm{1}}=\mathrm{120} \\ $$$$\mathrm{3}^{{n}} =\mathrm{1}+\mathrm{80}=\mathrm{81}=\mathrm{3}^{\mathrm{4}} \\ $$$$\Rightarrow{n}=\mathrm{4} \\ $$

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