`sum_(i=0)^oosum_(j=0)^oosum_(k=0)^oo1/(3^i3^i3^k)`

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Find the value of sum_(i=0)^(oo)sum_(j=0)^(oo)sum_(k=0)^(oo)(1)/(3^(i)3^(j)3^(k)) .

sum_(i=1)^(oo)sum_(j=1)^(oo)sum_(k=1)^(oo)(1)/(a^(i+j+k)) is equal to (where |a| gt 1 )

sum_(i=1)^(oo)sum_(j=1)^(oo)sum_(k=1)^(oo)(1)/(2^(i+j+k)) is equal to

sum_ (i = 0)^(oo) sum_ (j = 0)^(oo) sum_ (k = 0)^(oo) (1)/(3^(i) 3^(i) 3^(k) )

S=sum_(i=1)^(n)sum_(j=1)^(i)sum_(k=1)^(j)1

sum_(i=0)^(n)sum_(j=0)^(n)sum_(k=0)^(n)((n),(i))((n),(j))((n),(k)),((n),(r))=""^(n)C_(r) :

Value of sum_(k=1)^(oo)sum_(r=0)^(k)(1)/(3^(k))(^(k)C_(r)) is (2)/(3)b*(4)/(3)c.2d1

The value of sum_(i=1)^(n) sum_(j=1)^(i) sum_(k=1)^(j) 1 is

alpha=lim_(n rarr oo)sum_(i=1)^(n)sum_(j=1)^(i)(j)/(n^(3)), then [(1)/(alpha)-1] is