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Question Number 87933 by john santu last updated on 07/Apr/20

$$\mathrm{how}\mathrm{many}\mathrm{terms}\mathrm{are}\mathrm{there}$$$$\mathrm{in}{({\mathrm{x}}^2+\mathrm{x}+1)}^{11}?$$

Commented by mr W last updated on 07/Apr/20

$$x^0...x^{22}$$$$\Rightarrow 23terms$$

Commented by mr W last updated on 07/Apr/20

$$buthowmanytermsaretherein$$$${(x^5+x^2+1)}^{11}?$$

Commented by john santu last updated on 07/Apr/20

$$56\mathrm{terms}$$

Commented by john santu last updated on 07/Apr/20

$$\mathrm{hihihi}.....$$

Commented by mr W last updated on 07/Apr/20

$$butforexamplewe{don}^{\prime }thavexterm,$$$$x^{3}termetc.$$

Commented by jagoll last updated on 07/Apr/20

$$\underset{\mathrm{i}+\mathrm{j}+\mathrm{k}⩽11}{\sum ^{11}}\left(\begin{array}{c}11\\ \mathrm{i},\mathrm{j},\mathrm{k}\end{array}\right){\left({\mathrm{x}}^5\right)}^{\mathrm{i}}{\left({\mathrm{x}}^2\right)}^{\mathrm{j}}{\left(1\right)}^{\mathrm{k}}$$$$=\underset{\mathrm{i}+\mathrm{j}+\mathrm{k}⩽11}{\sum ^{11}}\left(\begin{array}{c}11\\ \mathrm{i},\mathrm{j},\mathrm{k}\end{array}\right){\mathrm{x}}^{5\mathrm{i}+2\mathrm{j}}{\left(1\right)}^{\mathrm{k}}$$$$\mathrm{i}=0\Rightarrow \mathrm{j}=0\wedge \mathrm{k}=11,$$$$\mathrm{j}=1\wedge \mathrm{k}=9,\mathrm{j}=2\wedge \mathrm{k}=7$$$$\mathrm{j}=3\wedge \mathrm{k}=5,\mathrm{j}=4\wedge \mathrm{k}=3$$$$\mathrm{j}=5\wedge \mathrm{k}=1$$$$\mathrm{i}=1\Rightarrow \mathrm{j}=0\wedge \mathrm{k}=6,\mathrm{j}=1\wedge \mathrm{k}=4$$$$\mathrm{j}=2\wedge \mathrm{k}=2,\mathrm{j}=3\wedge \mathrm{k}=0$$$$\mathrm{i}=2\Rightarrow \mathrm{j}=0\wedge \mathrm{k}=1$$$$\mathrm{\Sigma }\mathrm{terms}=11$$

Commented by jagoll last updated on 07/Apr/20

$$\mathrm{hehehe}$$

Commented by ajfour last updated on 07/Apr/20

$$Istillthink,mrWSirs,obvious$$$$answer,correct.$$

Commented by jagoll last updated on 08/Apr/20

$$\mathrm{what}\mathrm{answer}\mathrm{this}\mathrm{question}\mathrm{sir}?$$

Commented by mr W last updated on 08/Apr/20

$$whenexpanded,{(x^5+x^2+1)}^{11}has51$$$$terms.$$$$$$$${(x^5+x^2+1)}^{11}=\sum _{i+j+k=11}\left({}_{i,j,k}^{11}\right){\left(x^5\right)}^i{\left(x^2\right)}^j{1}^k$$$$=\sum _{i+j+k=11}\left({}_{i,j,k}^{11}\right)x^{5i+2j}$$$$i,j,k⩾0$$$$i+j+k=11\Rightarrow 0⩽i+j⩽11$$$$0⩽5i+2j⩽55$$$$butfrom0to55thereareonly51$$$$numberswhichisamultipleof5$$$$(max.55)oramultipleof2(max.22)$$$$orthesumofboth.$$$$$$$$followingterms{don}^{\prime }texist:$$$$x^1,x^3,x^{48},x^{51},x^{53}$$$$becauseyou{can}^{\prime }texpressthenumbers$$$$1,3,48,51,53as5i+2jwith0⩽i+j⩽11.$$

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