and=N+(2n 1) n-.1 +3+5+13. The identity(1+2+3++n)n 1= 13 +23 +33 + .+ n3,as the first century. Provide awas known as earlyderivation of it.14. Prove the following formula for the sum of triangularnumbers, given by the Hindu mathematicianAryabhata (circa 500):n(n +1)(n+2)t1t2 +t3 + +tnHint: Group the terms on the left-hand side in pairs,replacing t-1 + tk by k; consider the two cases wheren is odd and n is even.]15. Archimedes (287-212 B.C.) also derived the formula12+22+32 +n(n + 1)(2n + 1)+ n211for the sum of squares. Fill in any missing details in the HuSee 8-9-74ed to3714andSac 3.3 #CU9/20/19a denvahon of (1+2+3 n ++nHw 3.2# 133Pronden .(proof of it)t(+36इमई टाटterte sideiDt2 n13+23 3 n 1-1-1otos+n nn1-1-1 + 2 2-2l2-1+22+*inn+6P(I2t3+ n) n(nti):+ग(गर स) नाh(nt)(stnen tn42hCntl)(nt2)-- t+tnProve titt+t3+Hint: Group the terms in LHS in pairs, replacing(consider n is odd or even)K++Suppose n is even: (tt) (t3ttu)(tstto)tttntnJ!l

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i sort of got most of the way

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Step 1

To prove the required identity (for all n).

Step 2

Proof by mathematic...

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