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 a was known as early derivation of it. 14. Prove the following formula for the sum of triangular numbers, given by the Hindu mathematician Aryabhata (circa 500): n(n +1)(n+2) t1t2 +t3 + +tn Hint: Group the terms on the left-hand side in pairs, replacing t-1 + tk by k; consider the two cases where n is odd and n is even.] 15. Archimedes (287-212 B.C.) also derived the formula 12+22+32 + n(n + 1)(2n + 1) + n2 11 for the sum of squares. Fill in any missing details in the 4 Prove ttt+t3+ -- -tttn hCntl)(n+2) Hint: Group the terms in LHS in pais, replacing tx ttx wlK (csnsider n is odd or even) Suppose n is even: (tt) (t3+t) (tstto+tntn) 122232+ tn 2. 2. 4 2 ча 2 nin+1)12n+) Lo 2t4+ 2 (ak)2 + n ak k2 4C12+22+3 -- - + K2 4k(k+) (2k+1) Z igure us nint (nt2) la and smailar todd t n is odd t,ttztts+tytta tn-tth It3 +52+ t35 2422132142+5t+(k2+2K1)2 -(22+42t2(2k) ak+)ak+ 1} +1)(al2ki) 4K) akt S+D90 00rtsvnsts Sbwey nakt 2. +n2 2+ dd im forms missirg ble we ceunfaetsr anything out nintlnt) 1 11 I ( l

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 a was known as early derivation of it. 14. Prove the following formula for the sum of triangular numbers, given by the Hindu mathematician Aryabhata (circa 500): n(n +1)(n+2) t1t2 +t3 + +tn Hint: Group the terms on the left-hand side in pairs, replacing t-1 + tk by k; consider the two cases where n is odd and n is even.] 15. Archimedes (287-212 B.C.) also derived the formula 12+22+32 + n(n + 1)(2n + 1) + n2 11 for the sum of squares. Fill in any missing details in the 4 Prove ttt+t3+ -- -tttn hCntl)(n+2) Hint: Group the terms in LHS in pais, replacing tx ttx wlK (csnsider n is odd or even) Suppose n is even: (tt) (t3+t) (tstto+tntn) 122232+ tn 2. 2. 4 2 ча 2 nin+1)12n+) Lo 2t4+ 2 (ak)2 + n ak k2 4C12+22+3 -- - + K2 4k(k+) (2k+1) Z igure us nint (nt2) la and smailar todd t n is odd t,ttztts+tytta tn-tth It3 +52+ t35 2422132142+5t+(k2+2K1)2 -(22+42t2(2k) ak+)ak+ 1} +1)(al2ki) 4K) akt S+D90 00rtsvnsts Sbwey nakt 2. +n2 2+ dd im forms missirg ble we ceunfaetsr anything out nintlnt) 1 11 I ( l

Name: Number 14 please. Got partly through the problem and=N+(2n 1) n-.1 +3+
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Description: Number 14 please. Got partly through the problem 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 triangularnumbe

Algebra & Trigonometry with Analytic Geometry

13th Edition

ISBN:9781133382119

Author:Swokowski

Publisher:Swokowski

Chapter10: Sequences, Series, And Probability

Section10.4: Mathematical Induction

Problem 46E

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