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Question Number 181897 by Acem last updated on 01/Dec/22

Answered by mr W last updated on 02/Dec/22

T_k =(k^2 /(k^2 −10k+50))=(k^2 /((k−5)^2 +5^2 )) T_5 =(5^2 /5^2 )=1 T_1 +T_9 =(((5−4)^2 )/(4^2 +5^2 ))+(((5+4)^2 )/(4^2 +5^2 ))=((2(4^2 +5^2 ))/(4^2 +5^2 ))=2 T_2 +T_8 =(((5−3)^2 )/(3^2 +5^2 ))+(((5+3)^2 )/(3^2 +5^2 ))=((2(3^2 +5^2 ))/(3^2 +5^2 ))=2 T_3 +T_7 =...=2 T_4 +T_6 =...=2 ⇒T_1 +T_2 +...+T_9 =1+4×2=9 ✓

Tk=k2k2−10k+50=k2(k−5)2+52T5=5252=1T1+T9=(5−4)242+52+(5+4)242+52=2(42+52)42+52=2T2+T8=(5−3)232+52+(5+3)232+52=2(32+52)32+52=2T3+T7=...=2T4+T6=...=2⇒T1+T2+...+T9=1+4×2=9✓

Commented by Acem last updated on 02/Dec/22

Very well! Thanks

Verywell!Thanks

Answered by Acem last updated on 02/Dec/22

(1^2 /(1^2 −10+50))= ((2.1^2 )/(1^2 +1−20+100))= ((2.1^2 )/(1^2 + (10−1)^2 )) ...i (9^2 /(9^2 −90+50))= ((2 (10−1)^2 )/((10−1)^2 +(10−1)^2 −20(10−1)+100)) = ((2 (10−1)^2 )/((10−1)^2 +[10−(10−1)]^2 ))= ((2 (10−1)^2 )/(1^2 +(10−1)^2 )) ...ii i+ii = ((2 [1+(10−1)^2 ])/(1+(10−1)^2 ))= 2 Then ((2n^2 )/(n^2 +(10−n)^2 ))+ ((2 (10−n)^2 )/(n^2 +(10−n)^2 ))= 2 ∀n∈ N, n> 0 we have n= 9 : n=1 to 9 ; Terms= {1, 2, ..., 9} Sum= 4×2_(1,9& 2,8,...) + T_5 = 9 ; T_5 = 1

1212−10+50=2.1212+1−20+100=2.1212+(10−1)2...i9292−90+50=2(10−1)2(10−1)2+(10−1)2−20(10−1)+100=2(10−1)2(10−1)2+[10−(10−1)]2=2(10−1)212+(10−1)2...iii+ii=2[1+(10−1)2]1+(10−1)2=2Then2n2n2+(10−n)2+2(10−n)2n2+(10−n)2=2∀n∈N,n>0wehaven=9:n=1to9;Terms={1,2,...,9}Sum=4×21,9&2,8,...+T5=9;T5=1

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