Find the sum to n terms of the series 1/( 1 + 1^2 + 1^4) + 2/( 1 + 2^2 + 2^4) + 3/( 1 + 3^2 + 3^4) +......

urgent

The given series is, S= 11+12+14+21+22+24+31+32+34+.......upto n termsNow the nth term of the given series is, Tn=n1+n2+n4Now we know that , 

For x⁴+ x² + 1 

x⁴+ 1 + x² 

Add and subtract 2(x²)(1) to make a perfect square 

= (x²)² − 2(x²)(1) + 2(x²)(1) + (1)² + x² 

= (x² + 1)² − 2x² + x² 

= (x² + 1)² − x² 

= (x² + 1)² − (x)² 

Factorize as difference of 2 squares, a² − b² = (a + b)(a − b) 

= [ (x² + 1) + (x) ][ (x² + 1) − (x) ] 

= (x² + 1 + x)(x² + 1 − x) 

= (x² + x + 1)(x² − x + 1) 
Here x = n so we get, 
Tn= n1+n+n21-n+n2, now from here we have to adjust the term as follows, Tn=1211-n+n2-11+n+n2above two equation are same after simplifying second equation.Now T1=1211-13 T2=1213-17 T3=1217-113 T4=12113-121....Tn=1211-n+n2-11+n+n2Now after adding we can see that 13 will get cancelled with -13, similarly other terms tooand we will be left with first and last term so, Sn=T1+ T2+......+Tn=121-11+n+n2=12n+n21+n+n2

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