Evaluate sum r=1 ^ infty 1 r(r+2)^ prime

1^infinity = 1

Thus :

r = 1

r ( r + 2 )^prime =

1 × ( 1 + 2 )^1 =

1 × 3^1 =

1 × 3 =

3

.75 or 3/4

Step-by-step explanation:

One's that look sort of like this usually telescope when we write as partial fractions and expand a bit

So let's write as partial fractions

1/(r(r+2))=a/r+b/(r+2)

1=a(r+2)+br

If r=-2, then 1=b(-2) which implies b=-1/2

If r=0, then 1=a(2) which implies a=1/2

So 1/(r(r+2))=1/2(1/r-1/(r+2))

Let's expand the series a bit to see what's going on

1/2[1-1/3+1/2-1/4+1/3-1/5+1/4-1/6+1/5-1/7+1/6-1/8+1/7-1/9]

So everything in the bracket is "canceling" except 1+1/2 which equals 3/2

So we have

1/2[3/2]

3/4

0.75

the correct answer is 45x

9×5=45 then add the x

∞ this is your sign