calculate Σ_(n=1) ^(20) (1/n^2 )


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Question Number 74884 by abdomathmax last updated on 03/Dec/19

$c a l c u l a t e \sum_{n = 1}^{20} \frac{1}{n^{2}}$
Commented by mathmax by abdo last updated on 03/Dec/19
$let S=Σ_(n=1) ^(20) (1/n^2 ) ⇒ S =Σ_(p=1) ^(10) (1/((2p)^2 )) +Σ_(p=0) ^([((19)/2)]) (1/((2p+1)^2 )) =(1/4)Σ_(p=1) ^(10) (1/p^2 ) +Σ_(p=0) ^9 (1/((2p+1)^2 )) also Σ_(p=1) ^(10) (1/p^2 ) =(1/4)Σ_(p=1) ^5 (1/p^2 ) +Σ_(p=0) ^4 (1/((2p+1)^2 )) ⇒ S =(1/4){(1/4)Σ_(p=1) ^5 (1/p^2 ) +Σ_(p=0) ^4 (1/((2p+1)^2 ))}+Σ_(p=0) ^4 (1/((2p+1)^2 )) +Σ_(p=5) ^9 (1/((2p+1)^2 )) =(1/(16))Σ_(p=1) ^5 (1/p^2 ) +(5/4) Σ_(p=0) ^4 (1/((2p+1)^2 )) +Σ_(p=5) ^9 (1/((2p+1)^2 )) we have Σ_(p=5) ^9 (1/((2p+1)^2 )) =_(p−5=k) Σ_(k=0) ^4 (1/((2k+11)^2 )) ⇒ S =(1/(16))Σ_(p=1) ^5 (1/p^2 ) +(5/4)Σ_(p=0) ^4 (1/((2p+1)^2 )) +Σ_(p=0) ^4 (1/((2p+11)^2 )) =(1/(16))(1+(1/2^2 ) +(1/3^2 ) +(1/4^2 ) +(1/5^2 ))+(5/4)(1+(1/3^2 ) +(1/5^2 ) +(1/7^2 ) +(1/9^2 ) +(1/(11^2 ))) +(1/(11^2 )) +(1/(13^2 )) +(1/(15^2 )) +(1/(17^2 )) +(1/(19^2 )) S =(1/(16))(1+(1/4) +(1/9) +(1/(16)) +(1/(25)))+(5/4)(1+(1/9) +(1/(25)) +(1/(49)) +(1/(81)) +(1/(121))) +(1/(11^2 )) +(1/(13^2 )) +(1/(15^2 )) +(1/(17^2 )) +(1/(19^2 )) =....its eazy now to find S..$
$l e t S = \sum_{n = 1}^{20} \frac{1}{n^{2}} \Rightarrow S = \sum_{p = 1}^{10} \frac{1}{{(2 p)}^{2}} + \sum_{p = 0}^{[\frac{19}{2}]} \frac{1}{{(2 p + 1)}^{2}}$ $= \frac{1}{4} \sum_{p = 1}^{10} \frac{1}{p^{2}} + \sum_{p = 0}^{9} \frac{1}{{(2 p + 1)}^{2}} a l s o$ $\sum_{p = 1}^{10} \frac{1}{p^{2}} = \frac{1}{4} \sum_{p = 1}^{5} \frac{1}{p^{2}} + \sum_{p = 0}^{4} \frac{1}{{(2 p + 1)}^{2}} \Rightarrow$ $S = \frac{1}{4} {\frac{1}{4} \sum_{p = 1}^{5} \frac{1}{p^{2}} + \sum_{p = 0}^{4} \frac{1}{{(2 p + 1)}^{2}}} + \sum_{p = 0}^{4} \frac{1}{{(2 p + 1)}^{2}} + \sum_{p = 5}^{9} \frac{1}{{(2 p + 1)}^{2}}$ $= \frac{1}{16} \sum_{p = 1}^{5} \frac{1}{p^{2}} + \frac{5}{4} \sum_{p = 0}^{4} \frac{1}{{(2 p + 1)}^{2}} + \sum_{p = 5}^{9} \frac{1}{{(2 p + 1)}^{2}} w e h a v e$ $\sum_{p = 5}^{9} \frac{1}{{(2 p + 1)}^{2}} =_{p - 5 = k} \sum_{k = 0}^{4} \frac{1}{{(2 k + 11)}^{2}} \Rightarrow$ $S = \frac{1}{16} \sum_{p = 1}^{5} \frac{1}{p^{2}} + \frac{5}{4} \sum_{p = 0}^{4} \frac{1}{{(2 p + 1)}^{2}} + \sum_{p = 0}^{4} \frac{1}{{(2 p + 11)}^{2}}$ $= \frac{1}{16} (1 + \frac{1}{2^{2}} + \frac{1}{3^{2}} + \frac{1}{4^{2}} + \frac{1}{5^{2}}) + \frac{5}{4} (1 + \frac{1}{3^{2}} + \frac{1}{5^{2}} + \frac{1}{7^{2}} + \frac{1}{9^{2}} + \frac{1}{11^{2}})$ $+ \frac{1}{11^{2}} + \frac{1}{13^{2}} + \frac{1}{15^{2}} + \frac{1}{17^{2}} + \frac{1}{19^{2}}$ $S = \frac{1}{16} (1 + \frac{1}{4} + \frac{1}{9} + \frac{1}{16} + \frac{1}{25}) + \frac{5}{4} (1 + \frac{1}{9} + \frac{1}{25} + \frac{1}{49} + \frac{1}{81} + \frac{1}{121})$ $+ \frac{1}{11^{2}} + \frac{1}{13^{2}} + \frac{1}{15^{2}} + \frac{1}{17^{2}} + \frac{1}{19^{2}} = . . . . i t s e a z y n o w t o f i n d S . .$
Commented by mathmax by abdo last updated on 05/Dec/19
$another way let S =Σ_(n=1) ^(20) (1/n^2 ) ⇒ S = Σ_(n=1_(n=3k) ) ^(20) (1/n^2 ) +Σ_(n=1_(n=3k+1) ) ^(20) (1/n^2 ) +Σ_(n=1_(n=3k+2) ) ^(20) (1/n^2 ) =Σ_(n=1) ^([((20)/3)]) (1/(9n^2 )) +Σ_(n=0) ^([((19)/3)]) (1/((3k+1)^2 )) +Σ_(n=0) ^([((18)/3)]) (1/((3k+2)^2 )) =(1/9) Σ_(n=1) ^6 (1/n^2 ) +Σ_(n=0) ^6 (1/((3k+1)^2 )) +Σ_(n=0) ^6 (1/((3k+2)^2 )) =(1/9){ 1+(1/2^2 ) +(1/3^2 ) +(1/4^2 ) +(1/5^2 ) +(1/6^2 )} +{1+(1/4^2 ) +(1/7^2 ) +(1/(10^2 )) +(1/(13^2 )) +(1/(16^2 ))+(1/(19^2 ))} +{(1/2^2 ) +(1/5^2 ) +(1/8^2 ) +(1/(11^2 )) +(1/(14^2 )) +(1/(17^2 )) +(1/(20^2 ))} =....$
$a n o t h e r w a y l e t S = \sum_{n = 1}^{20} \frac{1}{n^{2}} \Rightarrow$ $S = \sum_{n = 1_{n = 3 k}}^{20} \frac{1}{n^{2}} + \sum_{n = 1_{n = 3 k + 1}}^{20} \frac{1}{n^{2}} + \sum_{n = 1_{n = 3 k + 2}}^{20} \frac{1}{n^{2}}$ $= \sum_{n = 1}^{[\frac{20}{3}]} \frac{1}{9 n^{2}} + \sum_{n = 0}^{[\frac{19}{3}]} \frac{1}{{(3 k + 1)}^{2}} + \sum_{n = 0}^{[\frac{18}{3}]} \frac{1}{{(3 k + 2)}^{2}}$ $= \frac{1}{9} \sum_{n = 1}^{6} \frac{1}{n^{2}} + \sum_{n = 0}^{6} \frac{1}{{(3 k + 1)}^{2}} + \sum_{n = 0}^{6} \frac{1}{{(3 k + 2)}^{2}}$ $= \frac{1}{9} {1 + \frac{1}{2^{2}} + \frac{1}{3^{2}} + \frac{1}{4^{2}} + \frac{1}{5^{2}} + \frac{1}{6^{2}}} + {1 + \frac{1}{4^{2}} + \frac{1}{7^{2}} + \frac{1}{10^{2}} + \frac{1}{13^{2}} + \frac{1}{16^{2}} + \frac{1}{19^{2}}}$ $+ {\frac{1}{2^{2}} + \frac{1}{5^{2}} + \frac{1}{8^{2}} + \frac{1}{11^{2}} + \frac{1}{14^{2}} + \frac{1}{17^{2}} + \frac{1}{20^{2}}} = . . . .$
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