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PROVE THAT 9*S2^2=S3(1+8*S1) where Sn is sum of n terms in arthematic progression.

Please give complete question.
the given question is a complete question.


The Brainliest Answer!
S_{1}→ Σn =  \frac{n(n+1)}{2}  .
S_{2} → Σn² =  \frac{n(n+1)(2n+1)}{6}  .
S_{3} → Σn³ =  \frac{n^{2}(n+1)^{2}}{4}  .
Consider, L.H.S. = (3S_{2})^{2}.
= (3\frac{n(n+1)(2n+1)}{6})^{2}
= (\frac{n(n+1)}{2})^{2}(2n+1)^{2}
= S_{3}(1+(4n^{2}+4n))
= S_{3}(1+4n(n+1))
= S_{3}(1+8 \frac{n(n+1)}{2})
= S_{3}(1+8S_{1}) = R.H.S.
2 5 2
thank u.
Sorry for the delay, it's because I'm using a tab an and the equation editor accidentally deleted the things when i was half way through.
not an issue
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