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Question Number 115908 by Eric002 last updated on 29/Sep/20

$$whatisthecofficientof{x}^2$$$$(1+x)(1+2x)(1+4x)......(1+2^{n}x)$$

Commented by soumyasaha last updated on 29/Sep/20

$$$$$$\mathrm{Coefficient}=$$$$\frac{1}{2}[{\left(\underset{\mathrm{r}=0}{\sum ^{\mathrm{n}}}{2}^{\mathrm{r}}\right)}^2-\underset{\mathrm{r}=0}{\sum ^{\mathrm{n}}}{\left(2^{\mathrm{r}}\right)}^2]$$$$$$$$=\frac{1}{2}[{(2^{\mathrm{n}+1}-1)}^2-\frac{1}{3}(4^{\mathrm{n}+1}-1)]$$$$$$

Commented by prakash jain last updated on 29/Sep/20

$$\mathrm{Thanks}$$

Answered by mr W last updated on 29/Sep/20

$$\frac{1}{2}\underset{k=0}{\sum ^n}{2}^k\underset{i=0\ne k}{\sum ^n}{2}^i$$$$=\frac{1}{2}\underset{k=0}{\sum ^n}{2}^k(\underset{i=0}{\sum ^n}{2}^i-2^k)$$$$=\frac{1}{2}\underset{k=0}{\sum ^n}{2}^k(\frac{2^{n+1}-1}{2-1}-2^k)$$$$=\frac{1}{2}\underset{k=0}{\sum ^n}{2}^k(2^{n+1}-1-2^k)$$$$=\frac{1}{2}(2^{n+1}-1)\underset{k=0}{\sum ^n}{2}^k-\frac{1}{2}\underset{k=0}{\sum ^n}{4}^k$$$$=\frac{1}{2}(2^{n+1}-1)(2^{n+1}-1)-\frac{1}{2}\times \frac{4^{n+1}-1}{4-1}$$$$=\frac{1}{2}[2^{2(n+1)}-2\times 2^{n+1}+1]-\frac{2^{2(n+1)}-1}{2\times 3}$$$$=\frac{2(2^{2n+1}+1)}{3}-2^{n+1}$$$$$$$$examplen=2:$$$$=\frac{2(2^5+1)}{3}-2^3=14$$

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