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@lalaly too:)

do you know standard formula for 1^2+2^2+3^2+.... n^2 =... ?

no :(

i mean how to expand?

Multiply the terms :)
\[(n)(n+1)(2n+1)\]

ok then i gt this
\[3n^2+2n^3+n\]

Divide this by 6,

ok then here it is \[\frac{n^2}{2}+\frac{n^3}{3}+\frac{n}{6}.\]

Obviously for positive n
\[\large \frac{n^3}{3}+\frac{n^2}{2}+\frac{n}{6} > \frac {n^3}{3} \]

yeah thanx:)

How do you conclude that it is “obvious” for a positive \(n\)? @ash2326

@abhyudaysingh12 got it or not?

for every 'n' it is true actually:)

How would you conclude that?

\(\large \frac{n^2}{2}+\frac{n}{6} >0\)
for n>0

^ that way

then add n^3/3 on both sides.

\[n^2 + n > 0 \iff n(n + 1)>0 \iff n +1>0\iff n>-1\]

BUT question says that n>0 so n>-1 is ignored
1,2,.........n
see @ParthKohli :)

But you asserted that it's true for ALL \(n\).

sorry for that statement
but that i gt in haste;)

lol okay

:D
good job;)

to discuss something is good job:)

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