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Question

The sum of series 1.2.4 + 2.3.5 + 3.4.6 + ----------- 20 terms.


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Solution

nth terms of the given series

tn = n(n+1)(n+3)

= n(n2 + 4n + 3)

= n3 + 4n2 + 3n

Sum to n terms sn = ni=1 tn

Sn = ni=1 (n3 + 4n2 + 3n) = ni=1n3 + 4ni=1n2 + 3ni=1n

= n2(n+1)24+4.(n)(n+1)(2n+1)6+3n(n+1)2

Sn =n(n+1)12 [3n(n + 1) + 8(2n + 1) + 18]

Sn =n(n+1)12 [3n2 + 19n + 26]

= n(n+1)(n+2)(3n+13)12 --------------------(1)

Substituting n = 20, in the equation 1

S20 = 20(21)(22)(73)12 = 56210


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