1 + ( 1 + x ) + ( 1 + x + x^2 ) + ( 1 + x + x^2 + x^3 ) + \cdots
let y = 1 + (1+ x) + (1+ x + x^2) + \cdots(1+x+x^2 + \cdots x^{n-1})
so y(1-x)=(1-x)+ 1(1-x^2)+ \cdots+(1-x^n)
= n-(x+x^2+ \cdots+ x^n)
= n-x(1+x+ \cdots+ x^{n-1})
= n-x\dfrac{1-x^n}{1-x}
or y = \dfrac{n}{1-x} - x \dfrac{1-x^n}{1-x^2}
the above ans is correct if x is not 1
if x = 1 then we have 1 + 2 + 3 + \cdots+n = \dfrac{n(n+1)}{2}
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