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Algebra Level 4

If the function $f$ satisfies the relation $f\left( x+y \right) = f\left( x \right) f\left( y \right)$ for all natural numbers $x,y$

And if $f\left( 1 \right) =2$ , and for a natural number $a$ , the summation below is satisfied.

$\large \displaystyle \sum _{ r=1 }^{ n }{ f\left( a+r \right) } =16\left( { 2 }^{ n }-1 \right)$

What is the value of $a$ ?

3 6 4 2

2 solutions

Aditya Jaisingh
Aug 1, 2014

given f(x+y)=f(x)f(y) f(1+1)=f(1)f(1)=2 2=2^2 f(3)=f(1+2)=f(1) f(2)=2 (2^2)=2^3 guessing the pattern f(something)=2^something... f(a+r)=2^(a+r) therefore, opening sigma we get 2^(a+1) + 2^(a+2) + 2^(a+3) +.............+2^(a+n)=2^4(2^n-1) taking 2^a common 2^a(2^1 + 2^2 + 2^3 +2^4 +.......+2^n)=2^4(2^n-1) the terms inside bracket forms GP (2^a) 2(2^n - 1)=2^4(2^n-1) 2^(a+1)*(2^n-1)=2^4(2^n-1) cancelling common term from LHS and RHS 2^(a+1)=2^(4) a+1=4 a=3............got the answer

marvellous

Kandarp Singh - 6 years, 10 months ago

Why $2^a * 2$ ?

Carlos David Nexans - 6 years, 10 months ago

sum of GP = a(r^n-1) where a=first term r=common ratio n=no. of terms here a=2 r=2 n=n hense sum=2(2^n-1) and already 2^a is common so
finally the exp comes out to be (2^a)*2

Aditya Jaisingh - 6 years, 10 months ago

Chew-Seong Cheong
Jan 25, 2015

$\begin{aligned} f(x+y) & = f(x)f(y) \\ f(1) & = 2 \\ \Rightarrow f(2) & = f(1+1) = 4 \\ \Rightarrow f(3) & = f(2+1) = 8 \\ \Rightarrow f(4) & = f(3+1) = 16 \\ ... f(n) & = 2^n \end{aligned}$

Now, we have:

$\displaystyle \begin{aligned} \sum _{r=1} ^n {f(a+r)} & = 16(2^n-1) \\ & = 16(1+2+4+...+2^n) \\ & = 16+32+64+...+2^{3+n} \\ & = f(4)+f(5)+f(6)+...+f(\boxed{3}+n) \\ \end{aligned}$

$\Rightarrow a = \boxed{3}$

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