Question Number 88000 by jdmath last updated on 07/Apr/20
$${Is}\:{there}\:{a}\:{formula}\:{to}\:{calculate}\: \\ $$$$ \\ $$$$\underset{{i}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{\mathrm{1}}{{i}^{\mathrm{2}} } \\ $$$${interms}\:{of}\:{n}..? \\ $$
Commented by abdomathmax last updated on 08/Apr/20
![i thnk?no?formulae but we can write Σ_(i=1) ^(n ) (1/i^2 ) =Σ_(i=3k) (...)+Σ_(i=3k+1) (...)+Σ_(i=3k+2) (...) =Σ_(k=1) ^([(n/3)]) (1/(9k^2 )) +Σ_(k=0) ^([((n−1)/3)]) (1/((3k+1)^2 )) +Σ_(k=0) ^([((n−2)/3)]) (1/((3k+2)^2 )) =....](https://www.tinkutara.com/question/Q88054.png)
$${i}\:{thnk}?{no}?{formulae}\:{but}\:{we}\:{can}\:{write} \\ $$$$\sum_{{i}=\mathrm{1}} ^{{n}\:} \:\frac{\mathrm{1}}{{i}^{\mathrm{2}} }\:=\sum_{{i}=\mathrm{3}{k}} \:\left(…\right)+\sum_{{i}=\mathrm{3}{k}+\mathrm{1}} \:\left(…\right)+\sum_{{i}=\mathrm{3}{k}+\mathrm{2}} \:\:\left(…\right) \\ $$$$=\sum_{{k}=\mathrm{1}} ^{\left[\frac{{n}}{\mathrm{3}}\right]} \:\frac{\mathrm{1}}{\mathrm{9}{k}^{\mathrm{2}} }\:+\sum_{{k}=\mathrm{0}} ^{\left[\frac{{n}−\mathrm{1}}{\mathrm{3}}\right]} \:\frac{\mathrm{1}}{\left(\mathrm{3}{k}+\mathrm{1}\right)^{\mathrm{2}} }\:+\sum_{{k}=\mathrm{0}} ^{\left[\frac{{n}−\mathrm{2}}{\mathrm{3}}\right]} \:\frac{\mathrm{1}}{\left(\mathrm{3}{k}+\mathrm{2}\right)^{\mathrm{2}} } \\ $$$$=…. \\ $$
Commented by jdmath last updated on 16/Apr/20
$${thank}\:{you}…{rly} \\ $$