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father

solve for 'n'.

_{[ k=1]}

^{[ n](} K^2+3K+3)[(k+1)!]=(2007*2007!)-4.

this is right question.

ramyadiamond

is it (2007*2007!)-4 or 2007 *(2007!-4)?

ramyadiamond

well, i'll tell u the procedure.

firstly distribute the sigma sign over all the terms,

and then apply formulae for the sum of first n terms and sum of squares of the first n terms.

k² +3

k +

3 =LHS

n(n+1)(2n+1)/6 + 3. n(n+1)/2 +3n =LHS

and on further simplifying u might get the answer

amar.gupta

dear,

ramya has shown you the procedure , try and check it out . If got some problem ,please feel free to ask.

and secondly i think, it should be

_{[ k=1]}

^{[ n](} K^2+3K+3) instead of _{[ r=1]}

^{[ n](} K^2+3K+3)

father

pleeeeeeeeeeeeeeez solve it quickly.

vinu

Hi father,

(k²+3k+3) = (k+2)(k+3)-2(k+2)+1

So, _[k=1]

^[n] (k²+3k+3) [(k+1)!] = _[k=1]^[n](k+3)! - 2_[k=1]^[n](k+2)! + _[k=1]^[n](k+1)!

= [4!+5!+...(n+3)!] -2 [3!+4!+...(n+2)!] + [2!+3!+....(n+1)!]

= [2!-3!-(n+2)!+(n+3)!]

=(n+2)(n+2)! - 4 ;

on comparing , u get (n+2) = 2007

or n = 2005.

elessar_iitkgp

Excellent solution Vinu

prasad_aspirant

yeah vinu has a great mind

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