IIT JEE , AIEEE Forum - Mathematics , Differential Calculus

deep01

find

_{[n ]}

_{[ ]} [1/(1-n²)+2/(1-n²)+-------------------------+n/1-n²]

pls also explain ur answer

vinod

Do correct me if something is wrong !

cheers ! !

iitkgp_bipin

S = _{[n ]}

_{[ ]} [1/(1-n²)+2/(1-n²)+-------------------------+n/1-n²]

= _{[n ]}

_{[ ]} (1+2+3+.....+n)/(1 - n²)

= _{[n ]}

_{[ ]} n(n+1)/2(1-n²)

= _{[n ]}

_{[ ]} (n²+n)/2(1-n²)

= _{[n ]}

_{[ ]} (1 + 1/n)/{2(1/n² - 1)}

= -1/2