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Question Number 22768 by math solver last updated on 22/Oct/17

Commented by math solver last updated on 22/Oct/17

find the sum of infinite terms of the series ..

$${find}\:{the}\:{sum}\:{of}\:{infinite}\:{terms}\:{of}\: \\ $$$${the}\:{series}\:.. \\ $$

Commented by math solver last updated on 22/Oct/17

someone solve this one plz .

$${someone}\:{solve}\:{this}\:{one}\:{plz}\:. \\ $$

Answered by $@ty@m last updated on 22/Oct/17

N^r of t_n =2n+1 D^r of t_n =1^2 +2^2 +...+n^2 =Σn^2 =((n(n+1)(2n+1))/6) (you can find its proof in your textbook) ∴ t_n =((6(2n+1))/(n(n+1)(2n+1))) =(6/(n(n+1))) =6((1/n)−(1/(n+1))) ∴S_n =6{(1−(1/2))+((1/2)−(1/3))+.....+((1/n)−(1/(n+1)))} =6(1−(1/(n+1))) =((6n)/(n+1)) S_∞ =lim_(n→∞) ((6n)/(n+1)) S_∞ =lim_(n→∞) (6/(1+(1/n)))=6 Ans.

$${N}^{{r}} {of}\:{t}_{{n}} =\mathrm{2}{n}+\mathrm{1} \\ $$$${D}^{{r}} {of}\:{t}_{{n}} =\mathrm{1}^{\mathrm{2}} +\mathrm{2}^{\mathrm{2}} +...+{n}^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\Sigma{n}^{\mathrm{2}} =\frac{{n}\left({n}+\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{1}\right)}{\mathrm{6}} \\ $$$$\left({you}\:{can}\:{find}\:{its}\:{proof}\:{in}\:{your}\:{textbook}\right) \\ $$$$ \\ $$$$\therefore\:{t}_{{n}} =\frac{\mathrm{6}\left(\mathrm{2}{n}+\mathrm{1}\right)}{{n}\left({n}+\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{1}\right)} \\ $$$$=\frac{\mathrm{6}}{{n}\left({n}+\mathrm{1}\right)} \\ $$$$=\mathrm{6}\left(\frac{\mathrm{1}}{{n}}−\frac{\mathrm{1}}{{n}+\mathrm{1}}\right) \\ $$$$\therefore{S}_{{n}} =\mathrm{6}\left\{\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}\right)+\left(\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{3}}\right)+.....+\left(\frac{\mathrm{1}}{{n}}−\frac{\mathrm{1}}{{n}+\mathrm{1}}\right)\right\} \\ $$$$=\mathrm{6}\left(\mathrm{1}−\frac{\mathrm{1}}{{n}+\mathrm{1}}\right) \\ $$$$=\frac{\mathrm{6}{n}}{{n}+\mathrm{1}} \\ $$$${S}_{\infty} =\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\:\:\frac{\mathrm{6}{n}}{{n}+\mathrm{1}} \\ $$$${S}_{\infty} =\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\:\:\frac{\mathrm{6}}{\mathrm{1}+\frac{\mathrm{1}}{{n}}}=\mathrm{6}\:{Ans}. \\ $$

Commented by math solver last updated on 22/Oct/17

how you write t_n directly ? plz tell steps ?? i always have problem finding that......

$${how}\:{you}\:{write}\:{t}_{{n}} \:{directly}\:? \\ $$$${plz}\:{tell}\:{steps}\:??\:{i}\:{always}\:{have}\:{problem} \\ $$$${finding}\:{that}...... \\ $$

Commented by $@ty@m last updated on 22/Oct/17

Is it clear now?

$${Is}\:{it}\:{clear}\:{now}? \\ $$

Commented by math solver last updated on 22/Oct/17

yes , thanks bro!

$${yes}\:\:,\:{thanks}\:{bro}! \\ $$

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