What is it simplified?
Thanks for the answer.

Question

Avery Alexander · Accepted Answer

# 3/sqrt10-1/sqrt2#.

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!

What is it simplified? Thanks for the answer.