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aakash.dixit · 1.3.5.7 + 3.5.7.9 + 5.7.9.11 + ............. n terms

1.3.5.7 + 3.5.7.9 + 5.7.9.11 + ............. n terms

aditya_arora04

general term : n(n+2)(n+4)(n+6)

open up and put the summation formulae for each term.

ultimostrikes

if u open that u'll get a term n⁴......wat's

n⁴???

sprinkle

Akash!

S = 1.3.5.7 + 3.5.7.9 + 5.7.9.11 + ............. n terms

=

(2r-1)(2r+1)(2r+3)(2r+5) where r = 1 to n

i.e. S =

Tr    where Tr = (2r-1)(2r+1)(2r+3)(2r+5)   and r = 1 to n

      Tr = (2r-1)(2r+1)(2r+3)(2r+5)

or   Tr = (1/10) [(2r-1)(2r+1)(2r+3)(2r+5)(2r+7) - (2r-3)(2r-1)(2r+1)(2r+3)(2r+5)]

=> T1 = (1/10) [ 1.3.5.7.9 - (-1).1.3.5.7]
     T2 = (1/10) [ 3.5.7.9.11 - 1.3.5.7.9]
     T3 = (1/10) [ 5.7.9.11.13 - 3.5.7.9.11]
     T4 = (1/10) [ 7.9.11.13.15 - 5.7.9.11.13]
.......................................................
    Tn = (1/10) [(2n-1)(2n+1)(2n+3)(2n+5)(2n+7) - (2n-3)(2n-1)(2n+1)(2n+3)(2n+5)]

(please note here that second part of every term is first part of previous term)

adding all of them,

S = T1 + T2 + T3 +......+ Tn

=>. S = (1/10) [(2n-1)(2n+1)(2n+3)(2n+5)(2n+7) - (-1).1.3.5.7]

=> S = (1/10) [(2n-1)(2n+1)(2n+3)(2n+5)(2n+7) + 105]         Ans

kap.mehra

sprinkle ji this is realy a amazing method.

sizzle_through

yes brilliant method!

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