Question

Please help me with this summation formula...

Answer 1

I only need help with part b.
I have already shown a and c but not quite sure what to do with b

Answer 2

can you share your work ? :)

Answer 3

i think a, b, c are steps in the proof

Answer 4

In step b I just solved for the sum of i^2, so all i need is to show that part b is true which will prove that the sum equals that formula

Answer 5

@ganeshie8

Answer 6

part a is algebra right?

Answer 7

To do so, I would recommend proving it by induction (since you seem to have proven the third step, already).

Answer 8

part b follows because you have a collapsing sum

Answer 9

no need for induction for this other than the implied induction for the collapsing sum

Answer 10

part c is also algebra

Answer 11

Well, yeah, it's just a straight up collapsing sum, by (a).

Answer 12

By collapsing sum do you mean converging ? I just haven't heard the term collapsing sum used before

Answer 13

I would recommend trying the first few cases numerically of (b) and seeing how it works :)

Answer 14

the whole point of the first part is this
\[(n+1)^2-n^2\] is one term minus the previous one

Answer 15

oops i meant
\[(n+1)^3-n^3\]

Answer 16

for example lets compute
\[\sum_{i=1}^5 (i+1)^3-i^3\] by just writing it all out

Answer 17

\[2^3-1^3+3^3-2^3+4^3-3^3+5^3-4^3=5^3-1\]

Answer 18

oops damn i missed the last term!!

Answer 19

\[\sum_{i=1}^5 (i+1)^3-i^3=2^3-1^3+3^3-2^3+4^3-3^3+5^3-4^3+6^3-5^3=6^3-1\]

Answer 20

which should convince you that
\[\sum_{i=1}^n(i+1)^3-i^3=(n+1)^3-1\]

Answer 21

and since by algebra \((i+1)^3-i^3=3i^2+3i+1\) that gives
\[(n+1)^2-1=\sum_{i=1}^n(3i^2+3i+1)\]

Answer 22

Thanks alot! and you forgot the cubed instead of squared again lol