calculate-n-1-20-1-n-2-

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

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}} = \dots . 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

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}}} = \dots .