( 2 n − 1 ) ( 1 ) + ( 2 n − 3 ) ( 2 ) + ( 2 n − 5 ) ( 3 ) + ( 2 n − 7 ) ( 4 ) + ⋯
What is the sum of all the positive terms of the expression above when n = 6 0 ?
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When n = 6 0 , the sum of positive terms becomes:
S = 1 1 9 ⋅ 1 + 1 1 7 ⋅ 2 + 1 1 5 ⋅ 3 + ⋯ + 1 ⋅ 6 0 = k = 1 ∑ 6 0 ( 1 2 1 − 2 k ) k = 1 2 1 k = 1 ∑ 6 0 k − 2 k = 1 ∑ 6 0 k 2 = 1 2 1 ⋅ 2 6 0 ⋅ 6 1 − 2 ⋅ 6 6 0 ⋅ 6 1 ⋅ 1 2 1 = 6 0 ⋅ 6 1 ⋅ 1 2 1 ( 2 1 − 3 1 ) = 1 0 ⋅ 6 1 ⋅ 1 2 1 = 7 3 8 1 0
T h e s e r i e s c a n b e w r i t t e n a s : 2 n − 1 2 n − 3 + 2 n − 3 2 n − 5 + 2 n − 5 + 2 n − 5 . . . . . . . . . 5 + 5 + 5 + . . . . . . + 5 n-2 times 3 + 3 + 3 + . . . . . . + 3 + 3 n-1 times 1 + 1 + 1 + . . . . . . + 1 + 1 + 1 n times n 2 + ( n − 1 ) 2 + ( n − 2 ) 2 + . . . . . . + 3 2 + 2 2 + 1 2 A d d i n g c o l u m n w i s e , w e g e t t h e s u m o f c o n s e c u t i v e o d d n u m b e r s i n e a c h c o l u m n . A s t h e s u m o f t h e f i r s t k o d d n u m b e r s i s k 2 , t h e p r o b l e m r e d u c e s t o f i n d i n g t h e s u m o f t h e f i r s t n s q u a r e s . H e n c e , t h e s u m i s ( n ( n + 1 ) ( 2 n + 1 ) ) / 6 . P l u g g i n g i n n = 6 0 , w e g e t 7 3 8 1 0 .
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Sum of { [2 × 60 – (2k – 1)] × k }
= Sum of { [121 – 2k] × k }
= Sum of { 121k – 2k² }
121 – 2k > 0, k < 60.5
k = [ 1 , 60 ]
Sum of { 121k – 2k² } for 1 ≤ k ≤ 60
= 121 × [k(k+1)/2] – 2 × [k(k+1)(2k+1)/6]
= k(k + 1) × [ 121/2 – (2k + 1)/3 ]
= 60 × 61 × [ 121/2 – 121/3 ]
= 73810