"Let f : R $\to$ R be a function which satisfies f(x + y) = f(x) + f(y) $\forall$ x, y $\in$ R. If f(1) = 2 and g(n) = $\sum\limits_{k = 1}^{\left( {n - 1} \right)} {f\left( k \right)}$, n $\in$ N then the value of n, for which g(n) = 20, is :"

Question

Accepted Answer

"Given f(1) = 2 ;

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

When x = y = 1 $\Rightarrow$ f(2) = 2 + 2 = 4

When x = 2, y = 1 $\Rightarrow$ f(3) = 4 + 2 = 6

g(n) = $\sum\limits_{k = 1}^{\left( {n - 1} \right)} {f\left( k \right)}$

= f(1) + f(2) +.........+ f(n - 1)

= 2 + 4 + 6 + ......+ 2(n - 1)

= 2 $\times$ $\frac{\left( n-1\right) \left( n\right) }{2}$

= n² - n

Given g(n) = 20

$\Rightarrow$ n²– n = 20

$\Rightarrow$ n² – n – 20 = 0

$\Rightarrow$ n = 5"

Let f : R $\to$ R be a function which satisfies
f(x + y) = f(x) + f(y) $\forall$ x, y $\in$ R. If f(1) = 2 and
g(n) = $\sum\limits_{k = 1}^{\left( {n - 1} \right)} {f\left( k \right)}$, n $\in$ N then the value of n, for which g(n) = 20, is :

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Let f : R $\to$ R be a function which satisfies f(x + y) = f(x) + f(y) $\forall$ x, y $\in$ R. If f(1) = 2 and g(n) = $\sum\limits_{k = 1}^{\left( {n - 1} \right)} {f\left( k \right)}$, n $\in$ N then the value of n, for which g(n) = 20, is :

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Drill 25 more like these. Every day. Free.

Let f : R $\to$ R be a function which satisfies
f(x + y) = f(x) + f(y) $\forall$ x, y $\in$ R. If f(1) = 2 and
g(n) = $\sum\limits_{k = 1}^{\left( {n - 1} \right)} {f\left( k \right)}$, n $\in$ N then the value of n, for which g(n) = 20, is :