Let for some R,f:R R be a function satisfying f(x+y)=f(x)+2y^2+y+ xy for all x,y R.

Question

Let for some $\alpha \in R,f:R	o R$ be a function satisfying $f(x+y)=f(x)+2{y}^{2}+y+\alpha xy$ for all $x,y\in R$. If $f(0)=-1$ and $f(1)=2$, then the value of $\sum _{n=1}^{5}(\alpha +f(n))$ is :

Accepted Answer

Given

$f(x+y)=f(x)+2{y}^{2}+y+\alpha xy\, x,y\in R$

with

$f(0)=-1,\, f(1)=2.$

We have to find

$\sum _{5}^{n=1}(\alpha +f(n)).$

Use the given relation to find $\alpha $: Put $x=0$ in

$f(x+y)=f(x)+2{y}^{2}+y+\alpha xy.$

Then

$f(y)=f(0)+2{y}^{2}+y+\alpha \cdot 0\cdot y$

so

$f(y)=-1+2{y}^{2}+y.$

Thus, for every real $y$,

$f(y)=2{y}^{2}+y-1.$

Now use $f(1)=2$:

$f(1)=2(1{)}^{2}+1-1=2.$

This is already satisfied, so now we must use the original equation to determine $\alpha $.

Substitute

$f(x+y)=2(x+y{)}^{2}+(x+y)-1$

and

$f(x)=2{x}^{2}+x-1$

into the given equation:

$2(x+y{)}^{2}+(x+y)-1=2{x}^{2}+x-1+2{y}^{2}+y+\alpha xy.$

Expand the left side:

$2{x}^{2}+4xy+2{y}^{2}+x+y-1.$

Right side is:

$2{x}^{2}+x-1+2{y}^{2}+y+\alpha xy.$

Comparing both sides,

$4xy=\alpha xy.$

Hence,

$\alpha =4.$

Find $f(n)$ for $n=1$ to $5$: Since

$f(n)=2{n}^{2}+n-1,$

we get:

$f(1)=2(1{)}^{2}+1-1=2$

$f(2)=2(2{)}^{2}+2-1=8+2-1=9$

$f(3)=2(3{)}^{2}+3-1=18+3-1=20$

$f(4)=2(4{)}^{2}+4-1=32+4-1=35$

$f(5)=2(5{)}^{2}+5-1=50+5-1=54$

Compute the required sum: We need

$\sum _{5}^{n=1}(\alpha +f(n)).$

Since $\alpha =4$,

$\alpha +f(1)=4+2=6$

$\alpha +f(2)=4+9=13$

$\alpha +f(3)=4+20=24$

$\alpha +f(4)=4+35=39$

$\alpha +f(5)=4+54=58$

Therefore,

$6+13+24+39+58=140.$

So the required value is

$140.$

Hence, the correct option is B.

Solutions · 1