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i-n-1-3i-1-2-3-n-1-i-0-




Question Number 6354 by sanusihammed last updated on 24/Jun/16
i = n−1  Σ  3i     =    (1/2)(3^n  − 1)  i = 0
i=n1Σ3i=12(3n1)i=0
Commented by nburiburu last updated on 24/Jun/16
doesn′t seem well writen:  Σ_(i=0) ^(n−1) 3i =3.Σ_(i=0) ^(n−1) i =3.n(n−1)/2   so in the way it is written it is FALSE.
doesntseemwellwriten:n1i=03i=3.n1i=0i=3.n(n1)/2sointhewayitiswrittenitisFALSE.
Commented by prakash jain last updated on 24/Jun/16
Σ_(i=0) ^(n−1) 3i≠(1/2)(3^n −1)  n=3  LHS=0+3+6=9  RHS=(1/2)(3^3 −1)=13  The formula is correct if the question is  Σ_(i=0) ^(n−1) 3^i   proving for n=k+1 assuming result for n=k  (1/2)(3^k −1)+=(1/2)[3^k −1+2∙3^k ]  =(1/3)[3^k (2+1)−1]=(1/2)(3^(k+1) −1)
n1i=03i12(3n1)n=3LHS=0+3+6=9RHS=12(331)=13Theformulaiscorrectifthequestionisn1i=03iprovingforn=k+1assumingresultforn=k12(3k1)+=12[3k1+23k]=13[3k(2+1)1]=12(3k+11)

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