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Question Number 7193 by peter james last updated on 15/Aug/16

Commented by Yozzia last updated on 15/Aug/16
$S(n)=1+2^2 −3^3 +4+5^2 −6^3 +7+8^2 −9^3 +... +[(3n−2)+(3n−1)^2 −(3n)^3 ] S(t=3n)=Σ_(k=1) ^n [(3k−2)+(3k−1)^2 −(3k)^3 ] In grouping the terms of S(n) in threes, t=3n is the total number of terms in the sum−expression of S(n). If we require 3n−1 terms, we can write S(t=3n−1)=Σ_(k=1) ^n [(3k−2)+(3k−1)^2 −(3k)^3 ]+27n^3 where n−1≡ −1 (mod 3) If we require 3n−2 terms, we write S(t=3n−2)=Σ_(k=1) ^n [(3k−2)+(3k−1)^2 −(3k)^3 ]+27n^3 −(3n−1)^2 where n−2≡ −2 (mod 3). So, S(t)= { ((S(t=3n) )),((S(t=3n−1)=S(t=3n)+27n^3 )),((S(t=3n−2)=S(t=3n)+27n^3 −(3n−1)^2 )) :} t=number of terms −−−−−−−−−−−−−−−−−−−−−−−−−− (3k−2)+(3k−1)^2 −27k^3 =3k−2+9k^2 −6k+1−27k^3 =9k^2 −1−3k−27k^3 Using : Σ_(k=1) ^n 1=n , Σ_(k=1) ^n k=((n(n+1))/2) Σ_(k=1) ^n k^2 =((n(n+1)(2n+1))/6) , Σ_(k=1) ^n k^3 =((n^2 (n+1)^2 )/4), S(n)=9×((n(n+1)(2n+1))/6)−n−((3n(n+1))/2)−((27n^2 (n+1)^2 )/4) S(n)=((3n(n+1)(2n+1))/2)−n−((3n(n+1))/2)−((27n^2 (n+1)^2 )/4) S(n)=(n/8)[12(n+1)(2n+1)−8−12(n+1)−54n(n+1)^2 ] S(n)=(n/8)[12(n+1)(2n+1−1)−8−54n(n+1)^2 ] S(n)=(n/8)[24n^2 +24n−8−54n^3 −108n^2 −54n] S(n)=(n/8)[−8−54n^3 −84n^2 −30n] S(n)=−(n/4)(27n^3 +42n^2 +15n+4) n=1⇒S(3)=((−1)/4)(27+42+15+4)=−22 (3 terms) n=2⇒ S(6)=((−1)/2)(27×8+42×4+15×2+4)=−209 (6 terms) t=5⇒S(5)=−209+6^3 =7=1+2^2 −3^3 +4+5^2 −−−−−−−−−−−−−−−−−−−−−−−− So, S(t)= { ((((−n)/4)(27n^3 +42n^2 +15n+4) t≡ 0 (mod 3) )),((((−n)/4)(27n^3 +42n^2 +15n+4)+27n^3 t≡ −1 (mod 3) )),((((−n(27n^3 +42n^2 +15n+4))/4)+27n^3 −(3n−1)^2 t≡ −2 (mod 3) )) :}$
$S (n) = 1 + 2^{2} - 3^{3} + 4 + 5^{2} - 6^{3} + 7 + 8^{2} - 9^{3} + . . .$ $+ [(3 n - 2) + {(3 n - 1)}^{2} - {(3 n)}^{3}]$ $S (t = 3 n) = \underset{k = 1}{\sum^{n}} [(3 k - 2) + {(3 k - 1)}^{2} - {(3 k)}^{3}]$ $I n g r o u p i n g t h e t e r m s o f S (n) i n t h r e e s,$ $t = 3 n i s t h e t o t a l n u m b e r o f t e r m s i n t h e$ $s u m - e x p r e s s i o n o f S (n) .$ $I f w e r e q u i r e 3 n - 1 t e r m s, w e c a n w r i t e$ $S (t = 3 n - 1) = \underset{k = 1}{\sum^{n}} [(3 k - 2) + {(3 k - 1)}^{2} - {(3 k)}^{3}] + 27 n^{3}$ $w h e r e n - 1 \equiv - 1 (m o d 3)$ $I f w e r e q u i r e 3 n - 2 t e r m s, w e w r i t e$ $S (t = 3 n - 2) = \underset{k = 1}{\sum^{n}} [(3 k - 2) + {(3 k - 1)}^{2} - {(3 k)}^{3}] + 27 n^{3} - {(3 n - 1)}^{2}$ $w h e r e n - 2 \equiv - 2 (m o d 3) .$ $S o, S (t) = {\begin{cases} S (t = 3 n) \\ S (t = 3 n - 1) = S (t = 3 n) + 27 n^{3} \\ S (t = 3 n - 2) = S (t = 3 n) + 27 n^{3} - {(3 n - 1)}^{2} \end{cases}$ $t = n u m b e r o f t e r m s$ $- - - - - - - - - - - - - - - - - - - - - - - - - -$ $(3 k - 2) + {(3 k - 1)}^{2} - 27 k^{3}$ $= 3 k - 2 + 9 k^{2} - 6 k + 1 - 27 k^{3}$ $= 9 k^{2} - 1 - 3 k - 27 k^{3}$ $U s i n g :$ $\underset{k = 1}{\sum^{n}} 1 = n, \underset{k = 1}{\sum^{n}} k = \frac{n (n + 1)}{2}$ $\underset{k = 1}{\sum^{n}} k^{2} = \frac{n (n + 1) (2 n + 1)}{6}, \underset{k = 1}{\sum^{n}} k^{3} = \frac{n^{2} {(n + 1)}^{2}}{4},$ $S (n) = 9 \times \frac{n (n + 1) (2 n + 1)}{6} - n - \frac{3 n (n + 1)}{2} - \frac{27 n^{2} {(n + 1)}^{2}}{4}$ $S (n) = \frac{3 n (n + 1) (2 n + 1)}{2} - n - \frac{3 n (n + 1)}{2} - \frac{27 n^{2} {(n + 1)}^{2}}{4}$ $S (n) = \frac{n}{8} [12 (n + 1) (2 n + 1) - 8 - 12 (n + 1) - 54 n {(n + 1)}^{2}]$ $S (n) = \frac{n}{8} [12 (n + 1) (2 n + 1 - 1) - 8 - 54 n {(n + 1)}^{2}]$ $S (n) = \frac{n}{8} [24 n^{2} + 24 n - 8 - 54 n^{3} - 108 n^{2} - 54 n]$ $S (n) = \frac{n}{8} [- 8 - 54 n^{3} - 84 n^{2} - 30 n]$ $S (n) = - \frac{n}{4} (27 n^{3} + 42 n^{2} + 15 n + 4)$ $n = 1 \Rightarrow S (3) = \frac{- 1}{4} (27 + 42 + 15 + 4) = - 22 (3 t e r m s)$ $n = 2 \Rightarrow S (6) = \frac{- 1}{2} (27 \times 8 + 42 \times 4 + 15 \times 2 + 4) = - 209 (6 t e r m s)$ $t = 5 \Rightarrow S (5) = - 209 + 6^{3} = 7 = 1 + 2^{2} - 3^{3} + 4 + 5^{2}$ $- - - - - - - - - - - - - - - - - - - - - - - -$ $S o, S (t) = {\begin{cases} \frac{- n}{4} (27 n^{3} + 42 n^{2} + 15 n + 4) t \equiv 0 (m o d 3) \\ \frac{- n}{4} (27 n^{3} + 42 n^{2} + 15 n + 4) + 27 n^{3} t \equiv - 1 (m o d 3) \\ \frac{- n (27 n^{3} + 42 n^{2} + 15 n + 4)}{4} + 27 n^{3} - {(3 n - 1)}^{2} t \equiv - 2 (m o d 3) \end{cases}$
Commented by peter james last updated on 16/Aug/16

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