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7.  If f is a function satisfying f (x +y) = f(x) f(y) for all x, y \epsilon N such that  f(1) = 3 and

        \sum_{x=1}^{n} f(x) = 120 , find the value of n.

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Given :  f (x +y) = f(x) f(y) for all x, y \epsilon N such that  f(1) = 3 

f(1) = 3

Taking x=y=1  , we have

f(1+1)=f(2)=f(1)*f(1)=3*3=9

f(1+1+1)=f(1+2)=f(1)*f(2)=3*9=27

f(1+1+1+1)=f(1+3)=f(1)*f(3)=3*27=81

f(1),f(2),f(3),f(4)..................... is 3,9,27,81,.............................. forms a GP with first term=3 and common ratio = 3.

 

\sum_{x=1}^{n} f(x) = 120=S_n

S_n=\frac{a(1-r^n)}{1-r}

120=\frac{3(1-3^n)}{1-3}

40=\frac{(1-3^n)}{-2}

-80=(1-3^n)

-80-1=(-3^n)

-81=(-3^n)

3^n=81

Therefore, n=4

Thus, value of n is 4.

Posted by

seema garhwal

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