sum-of-this-series-1-3-5-3-5-7-5-7-9-up-to-n-terms-

Question Number 40254 by scientist last updated on 17/Jul/18

s u m o f t h i s s e r i e s :

1.3.5 + 3.5.7 + 5.7.9 + \dots u p t o n t e r m s

Commented by maxmathsup by imad last updated on 18/Jul/18

\sum_{k = 0}^{n} (2 k + 1) (2 k + 3) (2 k + 5) = 1.3.5 + 3.5.7 + 5.7.9 + \dots . + (2 n + 1) (2 n + 3) (2 n + 5)

= S_{n}

S_{n} = \sum_{k = 0}^{n} (4 k^{2} + 6 k + 2 k + 3) (2 k + 5)

= \sum_{k = 0}^{n} (4 k^{2} + 8 k + 3) (2 k + 5)

= \sum_{k = 0}^{n} (8 k^{3} + 20 k^{2} + 16 k^{2} + 40 k + 6 k + 15)

= \sum_{k = 0}^{n} (8 k^{3} + 32 k^{2} + 46 k + 15)

= 8 \sum_{k = 0}^{n} k^{3} + 32 \sum_{k = 0}^{n} k^{2} + 46 \sum_{k = 0}^{n} k + 15 \sum_{k = 0}^{n} (1)

= 8 \frac{n^{2} {(n + 1)}^{2}}{4} + 32 \frac{n (n + 1) (2 n + 1)}{6} + 46 \frac{n (n + 1)}{2} + 15 n

= 2 n^{2} {(n + 1)}^{2} + \frac{16}{3} n (n + 1) (2 n + 1) + 23 n (n + 1) + 15 n .

Commented by scientist last updated on 18/Jul/18

b u t {c a n}^{'} t w e u s e T n = n (n + 2) (n + 4)

Commented by maxmathsup by imad last updated on 18/Jul/18

n o l o o k s i r \sum_{k = 1}^{n} T_{k} = \sum_{k = 1}^{n} k (k + 2) (k + 4) = 1.3.4 + 2.4.6 + \dots

a n d t h i s s u m i s t o t a l l y d i f f e r n t f r o m t h e s u m g i v e n i n t h e q u e s t i o n \dots