What is this equal to? 1+4+9+16+25...
Don't say that it is infinity.
d0=1222324252⋯1. Difference d1=357911⋯2. Difference d2=22222⋯
sn=(n1)⋅d0+(n2)⋅d1+(n3)⋅d2sn=(n1)⋅1+(n2)⋅3+(n3)⋅2(n1)=n(n2)=(n2)⋅(n−11)(n3)=(n3)⋅(n−12)⋅(n−21)sn=(n)⋅1+(n2)⋅(n−11)⋅3+(n3)⋅(n−12)⋅(n−21)⋅2|⋅66⋅sn=n⋅6+n⋅(n−1)⋅9+n⋅(n−1)⋅(n−2)⋅26⋅sn=n[ 6+(n−1)⋅9+(n−1)⋅(n−2)⋅2 ]6⋅sn=(n)[ 6+9n−9+(n2−3n+2)⋅2 ]6⋅sn=(n)[ −3+9n+(n2−3n+2)⋅2 ]6⋅sn=(n)( −3+9n+2n2−6n+4 )6⋅sn=(n)( 1+3n+2n2 )6⋅sn=(n)⋅(n+1)⋅(2n+1)sn=(n)⋅(n+1)⋅(2n+1)6s1=1=1⋅(1+1)⋅(2⋅1+1)6=1s2=1+4=2⋅(2+1)⋅(2⋅2+1)6=5s3=1+4+9=3⋅(3+1)⋅(2⋅3+1)6=14s4=1+4+9+16=4⋅(4+1)⋅(2⋅4+1)6=30s5=1+4+9+16+25=5⋅(5+1)⋅(2⋅5+1)6=55⋯
What is this equal to? 1+4+9+16+25...
Don't say that it is infinity.
d0=1222324252⋯1. Difference d1=357911⋯2. Difference d2=22222⋯
sn=(n1)⋅d0+(n2)⋅d1+(n3)⋅d2sn=(n1)⋅1+(n2)⋅3+(n3)⋅2(n1)=n(n2)=(n2)⋅(n−11)(n3)=(n3)⋅(n−12)⋅(n−21)sn=(n)⋅1+(n2)⋅(n−11)⋅3+(n3)⋅(n−12)⋅(n−21)⋅2|⋅66⋅sn=n⋅6+n⋅(n−1)⋅9+n⋅(n−1)⋅(n−2)⋅26⋅sn=n[ 6+(n−1)⋅9+(n−1)⋅(n−2)⋅2 ]6⋅sn=(n)[ 6+9n−9+(n2−3n+2)⋅2 ]6⋅sn=(n)[ −3+9n+(n2−3n+2)⋅2 ]6⋅sn=(n)( −3+9n+2n2−6n+4 )6⋅sn=(n)( 1+3n+2n2 )6⋅sn=(n)⋅(n+1)⋅(2n+1)sn=(n)⋅(n+1)⋅(2n+1)6s1=1=1⋅(1+1)⋅(2⋅1+1)6=1s2=1+4=2⋅(2+1)⋅(2⋅2+1)6=5s3=1+4+9=3⋅(3+1)⋅(2⋅3+1)6=14s4=1+4+9+16=4⋅(4+1)⋅(2⋅4+1)6=30s5=1+4+9+16+25=5⋅(5+1)⋅(2⋅5+1)6=55⋯
Thanks Heureka for this great answer.
I have just noticed that if I right click on your code I can bring it onto the screen in a resizable box.
I can also highlight it so I assume I can copy it and paste it somewhere else.
This is fabulous!
Mr Massow has been on the forum quite a bit recently. I assume this is one of the improvemtns he has made. :)
Thanks Mr Massow!! :D
What is this equal to? 1+4+9+16+25...
Don't say that it is infinity.
Hang on a moment.
If the number of terms is finite then Heureka's formula is great :/
But if the number of terms is not finite, which I think is implied by the dots, then lim as n approaches infinity is infinity