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Answer:

The given series is 3/(1^2*3^2) +5/( 2^2*3^2) +7/(3^2*4^2)...................19/(9^2*10^2)

Therefore  first term =( 2*1+1)/(1^2*3^2), second term =( 2*2+1)/( 2^2*3^2),

Third term = (2*3+1)/(3^2*4^2) and so the nth term =( 2*n+1)/{(n^2)*(n+1)^2}

So the last term=  19/(9^2*10^2) = (2*9+1)/ (9^2*10^2) = 9th term 

Let the first term of the given series = a1,second term =a2 and  

The sum of the first one term of the series=  S1,  a1 =  3/(1^1*2^2) =   3/4 

 Sum of the first two terms = S2= a1+a2 = 3/(1^2*2^2) +5/(2^2*3^2)  = (3/4)+5/(36) = 32/36= 8/9 

  Third term a3 = 7/(3^2*4^2)= 7/144    

Sum of the first three terms =  S3 = a1+a2+a3 = S2+a3 = (8/9) + (7/144) =  135/144=   15/16

Fourth term =  9/(4^2*5^2)=  9/400

Sum of the first four terms = a1+a2+a3+a4 =  S3+a4 = 15/16+(9/400)=    384/400= 24/25

 Now S1 = 3/4  ={(1+1)^2-1}/(1+1)^2.  ={(2^2)-1}/(2^2)

Similarly S2 = 8/9 = {(1+2)^2-1}/(1+2)^2 = {(3^2)-1}/3^2

S3 = 15/16 = {(4^2)-1)}/4^2  and S5 ={ (5^2)-1}/5^2  and so on.

 Therefore it can be seen that sum to n terms Sn = {(n+1)^2-1}/(n+1)^2

Now the last term in the given problem is = a9 = 19/(9^2*10^2)

Therefore S9=  a1+a2+a3+...............................a9 = {(10^2)-1}/(10^2) = 99/100

Therefore sum S9 =  99/100      

 

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