Question

F(x+y)=f(x).f(y) x,y€N f(1)=2 then sigma k=1 to 50 f(4+k)=?

Answer 1

The function $f$ is defined in such a way that,

$\longrightarrow f(x+y)=f(x)\cdot f(y)$

And given that $f(1)=2.$

Put $y=1.$ Then,

$\longrightarrow f(x+1)=f(x)\cdot f(1)$

$\longrightarrow f(x+1)=2f(x)$

This implies $(n+1)^{th}$ term of the sequence $f(1),\ f(2),\ f(3),\dots$ is twice its $n^{th}$ term.

Each term gets multiplied by 2 to obtain next term. And first term is $f(1)=2.$

So this sequence is a GP of first term $a=2$ and common ratio $r=2>1.$

$n^{th}$ term of this GP is,

$\longrightarrow f(n)=2\cdot2^{n-1}$

$\longrightarrow f(n)=2^n$

Sum of first $n$ terms of this GP is,

$\displaystyle\longrightarrow\sum_{k=1}^{n}f(k)=\dfrac{2(2^n-1)}{2-1}$

$\displaystyle\longrightarrow\sum_{k=1}^nf(k)=2(2^n-1)$

Therefore,

$\displaystyle\longrightarrow\sum_{k=1}^{50}f(4+k)=\sum_{k=5}^{54}f(k)$

$\displaystyle\longrightarrow\sum_{k=1}^{50}f(4+k)=\sum_{k=1}^{54}f(k)-\sum_{k=1}^{4}f(k)$

$\displaystyle\longrightarrow\sum_{k=1}^{50}f(4+k)=2(2^{54}-1)-2(2^4-1)$

$\displaystyle\longrightarrow\underline{\underline{\sum_{k=1}^{50}f(4+k)=2^{55}-2^5}}$

Answer 2

$\displaystyle\huge\red{\underline{\underline{Solution}}}$

FORMULA TO BE IMPLEMENTED

If in a Geometric Progression with first term = a and common ratio = r then the sum of first n terms is given by

$\displaystyle \: a \times \frac{ {r}^{n} - 1}{r - 1} \: \: \: \: \: \: \: \: where \: \: r > 1$

EVALUATION

It is given that

$f(x + y) = f(x)f(y)$

So

$f(2) = f(1 + 1) = f(1)(1) = 2 \times 2 = {2}^{2}$

$f(3) = f(2 + 1) = f(2)(1) = {2}^{2} \times 2 = {2}^{3}$

$f(4) = f(3 + 1) = f(3)(1) = {2}^{3} \times 2 = {2}^{4}$

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Similarly

$f(n) = {2}^{n}$

Now

$\displaystyle \sum\limits_{k=1}^{50} f(4 + k)$

$= f(5) + f(6) + f(7) + ....... + f(54)$

$= {2}^{5} + {2}^{6} + {2}^{7} + ..... + {2}^{54}$

This is a Geometric Progression with

$first \: term = {2}^{5} \: \: and \: common \: ratio \: = 2$

So

$\displaystyle \sum\limits_{k=1}^{50} f(4 + k)$

$\displaystyle \: = {2}^{5} \times \frac{ {2}^{50} - 1}{2 - 1}$

$\displaystyle \: = {2}^{5} \times ( {2}^{50} - 1 \: )$

$= {2}^{55} - {2}^{5}$