I would appreciate if somebody could help me with the following problem
Q: Find finite sum ? $$1^2\cdot 2^0+2^2\cdot 2^1+3^2\cdot 2^2+4^2\cdot 2^3+\cdots+n^2\cdot 2^{n-1}$$
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1Write $n^2 = n(n-1) + n$. Differentiate a geometric sum $x^0 + x^1 + \ldots x^k$ (twice). Check for similarities. – Daniel Fischer Jul 01 '13 at 11:40
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Do you know how to solve the slightly simpler problem of finding $$\sum_{k=1}^nkx^k;?$$ – Brian M. Scott Jul 01 '13 at 11:42
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$2^n ((n-2) n+3)-3$ – Ali Jul 01 '13 at 11:45
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1See the first answer here. (Under the "added note", in particular.) – David Mitra Jul 01 '13 at 11:48
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Hints (look at the comments, too):
$$|x|<1\implies \frac1{1-x}=\sum_{n=0}^\infty x^n\implies\frac1{(1-x)^2}=\sum_{n=1}^\infty nx^{n-1}\implies$$
$$\implies\frac2{(1-x)^3}=\sum_{n=2}^\infty n(n-1)x^{n-2}=\sum_{n=2}^\infty n^2x^{n-2} -\sum_{n=2}^\infty nx^{n-2}$$
Now multiply the whole second line above by $\,x\,$ and....well, and make some mathematics then.
DonAntonio
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$1^2\cdot 2^0+2^2\cdot 2^1+3^2\cdot 2^2+4^2\cdot 2^3+\cdots+n^2\cdot 2^{n-1}=S$-->1 multiply eq1 with $2$ and sustract from it which is, $$1^2.2^0+(2^2-1^2)2^1+......+(n^2-(n-1)^2).2^{n-1}-n^2.2^n=S$$ $$1^2.2^0+(2+1).2^1+(3+2).2^2+......+(n+(n-1).2^{n-1}-n^2.2^n=S$$ $$1.2^0+2.2^1+3.2^2+.......n.2^{n-1}+2(1.2^0+2.2^1+3.2^2+.......n.2^{n-1})-n^2.2^n-n.2.{n-1}=S$$ let,$$1.2^0+2.2^1+3.2^2+.......n.2^{n-1}=X$$ then,$$S=3X-n^2.2^n-n.2^{n-1}$$ you can calculate x if you know the technique asked by Brian.M.Scott in comments