Question

Here again who prove this without using mathematical induction method
1+2^2+3^2+4^2+⋯⋯+n^2=(n(n+1)(2n+1))/6

Answer 1

without mathematical induction ? >_<

Answer 2

yes

Answer 3

how is it possible ? >_<

Answer 4

Use concept of limit.

Answer 5

Oh sorry idk, probably someone else could help.

Answer 6

To be clear, the ONLY way to prove this in general is with the Principle of Mathematical Induction. There is no proof of this identity using the concept of limits.

Answer 7

Yes JamesJ it has.

Answer 8

Show us.

Answer 9

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

\[=\sum_{i=1}^{n}[3i^2+3i+1]=\sum_{i=1}^{n}3i^2+\sum_{i=1}^{n}3i+\sum_{i=1}^{n}1\]

\[=3\sum_{i=1}^{n}i^2+3\frac{n(n+1)}{2}+n\]

Solve for \[\sum_{i=1}^{n}i^2\]

Answer 10

Bhy is that first identity true:
\[ (n+1)^3 - 1 = \sum ... \]
and can you prove it without mathematical induction?

Note also the proof of \( \sum_1^n i = n(n+1)/2 \) also required induction.

Answer 11

the first identity is a telescoping sum



for the sum of i

s=1+2+...n
s=n+n-1+...+1
2s=(n+1)+...+(n+1)=n(n+1)
s=n(n+1)/2

Answer 12

Ok, fair enough then.

Answer 13

I stand corrected on the necessity of induction.

Answer 14

But I still call bs on the notion of using limits. :-)

Answer 15

cant win em all :)

my goal is at least 5% so that I can stay statistically relevant lol