We will attempt the following more General Solution :
#"Add : "(1-sqrt(1^2-1))/sqrt(1*2)+(2-sqrt(2^2-1))/sqrt(2.3)+(3-sqrt(3^2-1))/sqrt(3*4)+..."upto m terms"#.
Clearly, the General #n^(th)# Term, i.e., #T_n#, is given by,
#T_n=(n-sqrt(n^2-1))/sqrt(n(n+1))#,
#=n/sqrt(n(n+1))-sqrt(n^2-1)/sqrt(n(n+1))#.
#=(sqrtn*sqrtn)/(sqrtnsqrt(n+1))-(sqrt(n+1)sqrt(n-1))/(sqrtnsqrt(n+1))#,
#rArr T_n=sqrt(n/(n+1))-sqrt((n-1)/n)............................(ast)#.
#"Therefore, the sum "S_m=sum_(n=1)^(n=m)T_n#,
#=T_1+T_2+T_3+...+T_(m-1)+T_m#,
#={cancelsqrt(1/2)-sqrt(0/1)}+{cancelsqrt(2/3)-cancelsqrt(1/2)}+{cancelsqrt(3/4)-cancelsqrt(2/3)}+...+{cancelsqrt((m-1)/m)-cancelsqrt((m-2)/(m-1))}+{sqrt(m/(m+1))-cancelsqrt((m-1)/m)}...[because, (ast)]#,
#rArr S_m=sqrt(m/(m+1))-sqrt(0/1)=sqrt(m/(m+1))#.
Now, addressing to our case,
#"The Reqd. Sum="S_9-T_1=sqrt(9/10)-(1-sqrt(1^2-1))/sqrt(1*2)#,
#=sqrt(9/10)-1/sqrt2=3/sqrt10-1/sqrt2#.
Enjoy Maths., and Spread the Joy!
