Question

The natural number $\alpha$, for which $\underset{k=1}{\sum ^n}f(\alpha +k)=2n(31+n)$ where the function f satisfies the relation f(x+y)=f(x)+f(y) for all natural numbers x, y and further f(1)=4 is

• A. 10
• B. 15
• C. 16
• D. 20

Answer 1

The correct option is B 15

Given: $\underset{k=1}{\sum ^n}f(\alpha +k)=2n(31+n),f(x+y)=f\left(x\right)+f\left(y\right)$ and $f\left(1\right)=4$
Put n = 1
$f(\alpha +1)=2\left(1\right)(31+1)=64\Rightarrow f\left(\alpha \right)$ and $f\left(1\right)=64$
or $f\left(\alpha \right)=64-4=60\left(1\right)$
Put n = 2
$Putn=1:f(\alpha +2)=2\left(2\right)(31+2)=132f\left(\alpha \right)+f\left(1\right)+f\left(\alpha \right)+f\left(2\right)=1322f\left(\alpha \right)+f\left(1\right)+f\left(2\right)=1322.60+4+f\left(2\right)=132f\left(2\right)=8\left(2\right)$
Put n = 3
$f(\alpha +3)=2\left(3\right)(31+3)=2043f\left(\alpha \right)+f\left(1\right)+f\left(2\right)+f\left(3\right)=2043.60+4+8+f\left(3\right)=204\Rightarrow f\left(3\right)=12$
$f\left(1\right)=4;f\left(2\right)=8;f\left(3\right)=12\Rightarrow \left(3\right)$
$\Rightarrow f\left(\alpha \right)=60=4\alpha$
$\therefore \alpha =\frac{60}{4}=15$.