Given-only-the-standard-result-r-1-N-r-2-1-6-N-N-1-2N-1-is-applied-to-determining-the-series-1-2-2-2-2-3-2-2-4-2-5-2-2-6-2-7-2-2-n-1-2-n-2-where-n-is-odd-show-that-it-is-given

Question Number 669 by 112358 last updated on 21/Feb/15

G i v e n o n l y t h e s t a n d a r d r e s u l t \underset{r = 1}{\sum^{N}} r^{2} = \frac{1}{6} N (N + 1) (2 N + 1)

i s a p p l i e d t o d e t e r m i n i n g t h e s e r i e s

1^{2} + 2 \times 2^{2} + 3^{2} + 2 \times 4^{2} + 5^{2} + 2 \times 6^{2} + 7^{2} + \dots + 2 {(n - 1)}^{2} + n^{2}

w h e r e n i s o d d, s h o w t h a t i t i s g i v e n b y

t h e e x p r e s s i o n

\frac{1}{2} n^{2} (n + 1) .

Answered by 123456 last updated on 21/Feb/15

n i s o d d

n = 2 k + 1 \Leftrightarrow k = \frac{n - 1}{2}

S = 1^{2} + 2 \times 2^{2} + 3^{2} + 2 \times 4^{2} + \cdot \cdot \cdot + 2 \times {(2 k)}^{2} + {(2 k + 1)}^{2}

= [1^{2} + 3^{2} + \cdot \cdot \cdot + {(2 k + 1)}^{2}] + 2 [2^{2} + 4^{2} + \cdot \cdot \cdot + {(2 k)}^{2}]

= S_{1} + 2 S_{2}

S_{1} = 1^{2} + 3^{2} + \cdot \cdot \cdot + {(2 k + 1)}^{2} = \underset{r = 0}{\sum^{k}} {(2 r + 1)}^{2}

= \underset{r = 1}{\sum^{k + 1}} {[2 (r - 1) + 1]}^{2} = \underset{r = 1}{\sum^{k + 1}} {(2 r - 1)}^{2}

= \frac{1}{3} (k + 1) (2 k + 1) (2 k + 3)

S_{2} = 2^{2} + 4^{2} + \cdot \cdot \cdot + {(2 k)}^{2} = \underset{r = 1}{\sum^{k}} {(2 r)}^{2}

= 4 \underset{r = 1}{\sum^{k}} r^{2} = 4 \times \frac{1}{6} k (k + 1) (2 k + 1)

= \frac{2}{3} k (k + 1) (2 k + 1)

S = S_{1} + 2 S_{2}

= \frac{1}{3} (k + 1) (2 k + 1) (2 k + 3) + \frac{4}{3} k (k + 1) (2 k + 1)

= \frac{1}{3} (k + 1) (2 k + 1) (2 k + 3 + 4 k)

= \frac{1}{3} (k + 1) (2 k + 1) (6 k + 3)

= (k + 1) {(2 k + 1)}^{2}

= (k + 1) n^{2}

= (\frac{n - 1}{2} + 1) n^{2}

= \frac{1}{2} n^{2} (n + 1)

Commented by 123456 last updated on 21/Feb/15

S_{1} = \underset{r = 0}{\sum^{k}} {(2 r + 1)}^{2} = \underset{r = 0}{\sum^{k}} 4 r^{2} + 4 r + 1 = 1 + \underset{r = 1}{\sum^{k}} 4 r^{2} - 4 r + 1

S_{1} = \underset{r = 1}{\sum^{k + 1}} {(2 r - 1)}^{2} = \underset{r = 1}{\sum^{k + 1}} 4 r^{2} - 4 r + 1 = {(2 k + 1)}^{2} + \underset{r = 1}{\sum^{k}} 4 r^{2} - 4 r + 1

2 S_{1} = 1 + {(2 k + 1)}^{2} + \underset{r = 1}{\sum^{k}} 8 r^{2} + 2

2 S_{1} = 1 + 4 k^{2} + 4 k + 1 + 8 \underset{r = 1}{\sum^{k}} r^{2} + 2 \underset{r = 1}{\sum^{k}} 1

= 4 k^{2} + 4 k + 2 + 8 \times \frac{1}{6} k (k + 1) (2 k + 1) + 2 k

= 4 k (k + 1) + 2 + \frac{4}{3} k (k + 1) (2 k + 1) + 2 k

S_{1} = 2 k (k + 1) + \frac{2}{3} k (k + 1) (2 k + 1) + (k + 1)

= 2 k (k + 1) (1 + \frac{2 k + 1}{3}) + (k + 1)

= 2 k (k + 1) \frac{2 k + 4}{3} + (k + 1)

= (k + 1) [\frac{4 k (k + 2)}{3} + 1]

= (k + 1) \frac{4 k^{2} + 8 k + 3}{3}

= \frac{1}{3} (k + 1) (2 k + 1) (2 k + 3)

Commented by 112358 last updated on 21/Feb/15

T h a n k s

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