is-it-correct-when-S-n-1-2-1-3-1-4-1-5-1-n-2-3-2-4-2-5-2-n-3-4-3-5-3-n-4-5-4-n-

Question Number 8139 by sou1618 last updated on 01/Oct/16

i s i t c o r r e c t ?

w h e n

S_{n} = {

1 \times 2 + 1 \times 3 + 1 \times 4 + 1 \times 5 + \dots \dots . + 1 \times n

+ 2 \times 3 + 2 \times 4 + 2 \times 5 + \dots \dots . + 2 \times n

+ 3 \times 4 + 3 \times 5 + \dots \dots . + 3 \times n

+ 4 \times 5 + \dots \dots . + 4 \times n

\dots .

+ (n - 1) \times n

}

f i n d S_{n} .

/ / / / / / / / / / / / / / /

S_{n} + S_{n} + (1^{2} + 2^{2} + 3^{2} + 4^{2} + \dots + n^{2}) = {

1 \times 1 + 1 \times 2 + 1 \times 3 + 1 \times 4 + \dots \dots . + 1 \times n +

2 \times 1 + 2 \times 2 + 2 \times 3 + 2 \times 4 + \dots \dots . + 2 \times n +

3 \times 1 + 3 \times 2 + 3 \times 3 + 3 \times 4 + \dots \dots . + 3 \times n +

4 \times 1 + 4 \times 1 + 4 \times 3 + 4 \times 4 + \dots \dots . + 4 \times n +

\dots

n \times 1 + n \times 2 + n \times 3 + n \times 4 + \dots \dots + n \times n}

\Leftrightarrow

= (1 + 2 + 3 + 4 + \dots + n) (1 + 2 + 3 + 4 + \dots + n)

\Leftrightarrow

2 S_{n} + \frac{n (n + 1) (2 n + 1)}{6} = {\frac{n (n + 1)}{2}}^{2}

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

= \frac{n (n + 1)}{2} \times \frac{3 n^{2} - n - 2}{6}

S_{n} = \frac{n (n + 1)}{4} \times \frac{(3 n + 2) (n - 1)}{6}

S_{n} = \frac{(n - 1) n (n + 1) (3 n + 2)}{24}

Commented by prakash jain last updated on 01/Oct/16

This is correct . I will update my answer .

and check for mistakes .

Commented by prakash jain last updated on 01/Oct/16

Thanks . I have also corrected my answer .

I had started wrong .

The sum is \underset{j = 1}{\sum^{n - 1}} [\underset{i = j + 1}{\sum^{n}} j \cdot i]

earlier i had started wrongly

\underset{j = 1}{\sum^{n - 1}} [\underset{i = j + 1}{\sum^{n}} i (i + 1)]

Commented by sou1618 last updated on 02/Oct/16

t h a n k s!

Commented by 314159 last updated on 02/Oct/16

t h a n k s!

Answered by prakash jain last updated on 02/Oct/16

answer in question