1000-2-999-2-998-2-997-2-2-2-1-2-Please-the-question-says-simplify-

Question Number 6885 by Tawakalitu. last updated on 01/Aug/16

{(1000)}^{2} - {(999)}^{2} + {(998)}^{2} - {(997)}^{2} + \dots \dots \dots + 2^{2} - 1^{2} = ?

P l e a s e t h e q u e s t i o n s a y s s i m p l i f y

Answered by sou1618 last updated on 01/Aug/16

S = (1000^{2} - 999^{2}) + \dots + {{(2 k)}^{2} - {(2 k - 1)}^{2}} + \dots + (2^{2} - 1^{2})

1 <= k <= 500

S = (1000 + 999) (1000 - 999) + \dots + (2 k + 2 k - 1) (2 k - 2 k + 1) + \dots + (2 + 1) (2 - 1)

S = 1999 \times 1 + \dots + (4 k - 1) \times 1 + \dots + 3 \times 1

S = 1999 + 1995 + \dots + (4 k - 1) + \dots + 7 + 3

(i)

S = \underset{k = 1}{\sum^{500}} (4 k - 1)

S = 2 \times 500 \times (500 + 1) - 500

S = 501000 - 500

S = 500500

(i i)

S = (1999 + 3) + (1995 + 7) + \dots \dots + (1003 + 999)

{\begin{cases} 1003 = 4 \times 251 - 1 \\ 999 = 4 \times 250 - 1 \end{cases}

S = 2002 + 2002 + \dots . + 2002 (250 t i m e s)

{\begin{cases} 1999 = 4 \times 500 - 1 \\ 1003 = 4 \times 251 - 1 \end{cases}

S = 2002 \times 250

S = 500500

Commented by sou1618 last updated on 01/Aug/16

(i i i) o t h e r w a y

S = \underset{k = 1}{\sum^{500}} {(2 k)}^{2} - \underset{k = 1}{\sum^{500}} {(2 k - 1)}^{2}

S = \underset{k = 1}{\sum^{500}} {4 k^{2} - (4 k^{2} - 4 k + 1)}

S = \underset{k = 1}{\sum^{500}} (4 k - 1)

S = 2 \times 500 \times 501 - 500

S = 500500

Commented by Tawakalitu. last updated on 01/Aug/16

W o w t h a n k s s o m u c h

Commented by Tawakalitu. last updated on 01/Aug/16

I r e a l l y a p p r e c i a t e