Question 71405: sum the series
1+4/6 + 4.5/6.9 + 4.5.6/6.9.12 + ................infinity Found 2 solutions by Smirnov, hafys14feb:Answer by Smirnov(15) (Show Source):
You can put this solution on YOUR website! 1+ 4/6+ 4.5/6.9+ 4.5.6/6.9.12+ ...infinity
Let S= 1+ 4/6+ 4.5/6.9+ 4.5.6/6.9.12+ ...infinity
= 1+ 4/2.3 + 4.5/2.3.3.3 + 4.5.6/2.3.3.3.3.4 +...
= 1+ 4/1.2 [1/3] + 4.5/1.2.3 [1/3]^2 + 4.5.6/1.2.3.4 [1/3]^3 +...
Inserting the missing factor in the numerator
Multiply both sides by 3 ....(4,5,6 ...are in Ap.the preceding term is 3)
3S =3+ 3.4/2![1/3] +3.4.5/3! [1/3]^2 +3.4.5.6/4! [1/3]^3+...
Now Multiply both sides by [1/3]..
3S= 3/1! [1/3]+ 3.4/2! [1/3]^2 + 3.4.5/3! [1/3]^3 + ...
Now add 1 on both sides ;
S+1=1+3[1/3]+ 3.4/2![1/3]^2 + 3.4.5/3![1/3]^3 +...
This is in the form of
1+p/1![x/q] +p(p+q)/2![x/q]^2+p(p+q)(p+2q)/3! [x/q]^3...
which is equal to (1-x)^-p/q
So ,
S+1= (1-x)^-p/q ______equation 1
here p=3 and p+q=4
Therefore q=4-3=1
and we know x/q=1/3
so x= 1/3.
Again by solving the equation 1,
S+1= (1-1/3)^-3/1
=(2/3)^-3
S = 19/8
The sum is 19/8.
(
