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

Answer:

Given:–

• The equation is,f(x+y)=f(x)+f(y)
• And f(1)=2

To find:–

• sigma k=1 to 50 f(4+k)=?

Solution:–

As we know that,

The condition of the equation,

f(x+y)=f(x).f(y) is

$f(x) = {a}^{x}$

By using this condition,

Put,x=1

$f(1) = {a}^{1}$

Since,f(1)=2

${a}^{1} = 2$

$a = 2$

So,The value of a is 2.

Now,

By using the condition,

$f(x) = {a}^{x}$

put,x=2

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

$f(2) = 2²=4$

Again,

put x=3

$f(3) = {a}^{3}$

$f(3) = {2}^{3}$

Like this,

f(4)=2⁴,

f(5)=2⁵

f(6)=2⁶......

Now,

→ sigma k=1 to 50 f(4+k)=f(5)+f(6)+f(7)+......+f(54)

→ 2⁵+2⁶+2⁷+......+2⁵⁴

It is in the form of GP

→ First term=2⁵,

Common ratio,r=2

Now,

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

$\frac{ {2}^{5}( {2}^{5</strong><strong>0</strong><strong>} - 1) }{2 - 1}$

${2}^{5} ( {2}^{50} - 1)$

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

Additional information:–

The conditions for the following equations are,

• $f(x + y) = f(x).f(y)$→ $f(x) = {a}^{x}$
• $f(x + y) = f(x) + f(y) \\ f(x) = kx$
• $f(xy) = f(x).f(y) \\ f(x) = {x}^{n}$
• $f(xy) = f(x) + f(y) \\ f(x) = k log(x)$