IIT JEE , AIEEE Forum - Mathematics , Algebra

scadez

if

_{[ r=1]}

^{[ n]} t_{r ^{=n(n+1)(n+2)(n+3)/12 ,find}[r=1 ]}

^{[ infinity]} 1/tr........

scadez

itz one of my fav questions........its pretty long.....surely i will rate who ever solves it 1st(experts not included)

Avinash_Bhat

_{[ r=1]}

^{[ n]} t_{r ^{= n( n + 1 )( n + 2 )( n + 3 ) / 12

n( n + 1 )( n + 2 )( n + 3 ) / 12   =   ( n}}⁴_{^{+ 6n}}³_{^{+ 11n}}²_{^{+ 6n ) / 12

   =   [ { n( n + 1 ) / 2 }}}²_{^{+ 3{ n( n + 1 )( 2n + 1 ) / 6 } + 2{ n( n + 1 ) / 2 } ] / 3

   = {}}

_⁽ⁿ³_{^{) + 3}}

_⁽ⁿ²_{^{) + 2}}

_{^{(n) } / 3 =}}_{[ r=1]}

^{[ n]} t_r_{^{SO : t}r ^{= ( n}}³_{^{+ 3n}}²_{^{+ 2n ) / 3 = n( n}}²_{^{+ 3n + 2) / 3}}_{^{: 1 / t}r^{= 3 / n( n + 1)( n + 2 ) = 3[ { 1 / 2n } - { 1 / ( n + 1 ) } + { 1 / 2(n + 2) } ]

SO :}}_[r=1]

^[infinity]_{^{{ 1 / t}r^{} =   3{ [ 1/2 + 1/4 + 1/6 + 1/8 + 1/10 + ............... ( upto infinity ) ] -
[ 1/2 + 1/3 + 1/4 + 1/5 + 1/6 + 1/7 + 1/8 + ................ ( upto infinity ) ] +

[ 1/6 + 1/8 + 1/10 + 1/12 + 1/14 + 1/16 ............... ( upto infinity ) ]   }

   = 3{ - 1/3 - 1/5 + 1/6 - 1/7 + 1/8 - 1/9 + 1/10 - ................ ( upto infinity )}

BUT : - log 2   = - 1 + 1/2 - 1/3 + 1/4 - 1/5 + 1/6 - 1/ + 1/8 - ................ ( upto infinity )

SO :}}3{ - 1/3 - 1/5 + 1/6 - 1/7 + 1/8 - 1/9 + 1/10 - ................ ( upto infinity )} = 3 { - log 2 + 1/4 }

                                           =   log { e^3/4 / 8 }

scadez

answer is only 3/4

Avinash_Bhat

PLEASE SITE THE MISTAKE DONE............

AND.............GIVE THE STEPS FOR YOUR ANSWER.............

scadez

http://www.goiit.com/posts/list/algebra-rise-up-to-the-challenge-29406.htm

me

we r given sum of series ,let us denote it by Sn

now tn = Sn-Sn-1 = n(n+1)(n+2)/3

hence we get 1/tn = 3/n(n+1)(n+2) = 3[1/(n+1)n - 1/(n+1)(n+2)] / 2

hence 1/tn =3/2 [1/n(n+1) - 1/(n+1)(n+2)]

put n=1,2,3,4,ect. & add to get

[r=1 ][ infinity] 1/tr = 3/2[1/2 - 1/(n+1)(n+2)] = 3/4 {n tends to infinity}

Ask & Discuss Questions with Community & Experts