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

(1000)^2 − (999)^2 + (998)^2 − (997)^2 + ......... + 2^2 − 1^2 = ? Please the question says simplify

$$\left(\mathrm{1000}\right)^{\mathrm{2}} \:−\:\left(\mathrm{999}\right)^{\mathrm{2}} \:+\:\left(\mathrm{998}\right)^{\mathrm{2}} \:−\:\left(\mathrm{997}\right)^{\mathrm{2}} +\:.........\:+\:\mathrm{2}^{\mathrm{2}} \:−\:\mathrm{1}^{\mathrm{2}} \:=\:? \\ $$$$ \\ $$$${Please}\:{the}\:{question}\:{says}\:{simplify} \\ $$

Answered by sou1618 last updated on 01/Aug/16

S=(1000^2 −999^2 )+...+{(2k)^2 −(2k−1)^2 }+...+(2^2 −1^2 ) 1<=k<=500 S=(1000+999)(1000−999)+...+(2k+2k−1)(2k−2k+1)+...+(2+1)(2−1) S=1999×1+...+(4k−1)×1+...+3×1 S=1999+1995+...+(4k−1)+...+7+3 (i) S=Σ_(k=1) ^(500) (4k−1) S=2×500×(500+1)−500 S=501000−500 S=500500 (ii) S=(1999+3)+(1995+7)+......+(1003+999) { ((1003=4×251−1)),((999=4×250−1)) :} S=2002+2002+....+2002(250times) { ((1999=4×500−1)),((1003=4×251−1)) :} S=2002×250 S=500500

$${S}=\left(\mathrm{1000}^{\mathrm{2}} −\mathrm{999}^{\mathrm{2}} \right)+...+\left\{\left(\mathrm{2}{k}\right)^{\mathrm{2}} −\left(\mathrm{2}{k}−\mathrm{1}\right)^{\mathrm{2}} \right\}+...+\left(\mathrm{2}^{\mathrm{2}} −\mathrm{1}^{\mathrm{2}} \right) \\ $$$$\mathrm{1}<={k}<=\mathrm{500} \\ $$$${S}=\left(\mathrm{1000}+\mathrm{999}\right)\left(\mathrm{1000}−\mathrm{999}\right)+...+\left(\mathrm{2}{k}+\mathrm{2}{k}−\mathrm{1}\right)\left(\mathrm{2}{k}−\mathrm{2}{k}+\mathrm{1}\right)+...+\left(\mathrm{2}+\mathrm{1}\right)\left(\mathrm{2}−\mathrm{1}\right) \\ $$$${S}=\mathrm{1999}×\mathrm{1}+...+\left(\mathrm{4}{k}−\mathrm{1}\right)×\mathrm{1}+...+\mathrm{3}×\mathrm{1} \\ $$$${S}=\mathrm{1999}+\mathrm{1995}+...+\left(\mathrm{4}{k}−\mathrm{1}\right)+...+\mathrm{7}+\mathrm{3} \\ $$$$\left({i}\right) \\ $$$${S}=\underset{{k}=\mathrm{1}} {\overset{\mathrm{500}} {\sum}}\left(\mathrm{4}{k}−\mathrm{1}\right) \\ $$$${S}=\mathrm{2}×\mathrm{500}×\left(\mathrm{500}+\mathrm{1}\right)−\mathrm{500} \\ $$$${S}=\mathrm{501000}−\mathrm{500} \\ $$$${S}=\mathrm{500500} \\ $$$$ \\ $$$$\left({ii}\right) \\ $$$${S}=\left(\mathrm{1999}+\mathrm{3}\right)+\left(\mathrm{1995}+\mathrm{7}\right)+......+\left(\mathrm{1003}+\mathrm{999}\right) \\ $$$$\begin{cases}{\mathrm{1003}=\mathrm{4}×\mathrm{251}−\mathrm{1}}\\{\mathrm{999}=\mathrm{4}×\mathrm{250}−\mathrm{1}}\end{cases} \\ $$$${S}=\mathrm{2002}+\mathrm{2002}+....+\mathrm{2002}\left(\mathrm{250}{times}\right) \\ $$$$\begin{cases}{\mathrm{1999}=\mathrm{4}×\mathrm{500}−\mathrm{1}}\\{\mathrm{1003}=\mathrm{4}×\mathrm{251}−\mathrm{1}}\end{cases} \\ $$$${S}=\mathrm{2002}×\mathrm{250} \\ $$$${S}=\mathrm{500500} \\ $$

Commented by sou1618 last updated on 01/Aug/16

(iii)other way S=Σ_(k=1) ^(500) (2k)^2 −Σ_(k=1) ^(500) (2k−1)^2 S=Σ_(k=1) ^(500) {4k^2 −(4k^2 −4k+1)} S=Σ_(k=1) ^(500) (4k−1) S=2×500×501−500 S=500500

$$\left({iii}\right){other}\:{way} \\ $$$${S}=\underset{{k}=\mathrm{1}} {\overset{\mathrm{500}} {\sum}}\left(\mathrm{2}{k}\right)^{\mathrm{2}} −\underset{{k}=\mathrm{1}} {\overset{\mathrm{500}} {\sum}}\left(\mathrm{2}{k}−\mathrm{1}\right)^{\mathrm{2}} \\ $$$${S}=\underset{{k}=\mathrm{1}} {\overset{\mathrm{500}} {\sum}}\left\{\mathrm{4}{k}^{\mathrm{2}} −\left(\mathrm{4}{k}^{\mathrm{2}} −\mathrm{4}{k}+\mathrm{1}\right)\right\} \\ $$$${S}=\underset{{k}=\mathrm{1}} {\overset{\mathrm{500}} {\sum}}\left(\mathrm{4}{k}−\mathrm{1}\right) \\ $$$${S}=\mathrm{2}×\mathrm{500}×\mathrm{501}−\mathrm{500} \\ $$$${S}=\mathrm{500500} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$

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

Wow thanks so much

$${Wow}\:{thanks}\:{so}\:{much} \\ $$

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

I really appreciate

$${I}\:{really}\:{appreciate} \\ $$

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