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

how many terms are there in (x^2 +x+1)^(11) ?

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

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

x^0 ... x^(22) ⇒23 terms

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

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

but how many terms are there in (x^5 +x^2 +1)^(11) ?

$${but}\:{how}\:{many}\:{terms}\:{are}\:{there}\:{in} \\ $$$$\left({x}^{\mathrm{5}} +{x}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{11}} \:? \\ $$

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

56 terms

$$\mathrm{56}\:\mathrm{terms} \\ $$

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

hihihi.....

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

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

but for example we don′t have x term, x^3 term etc.

$${but}\:{for}\:{example}\:{we}\:{don}'{t}\:{have}\:{x}\:{term}, \\ $$$${x}^{\mathrm{3}} \:{term}\:{etc}. \\ $$

Commented by jagoll last updated on 07/Apr/20

Σ_(i+j+k ≤11) ^(11) ((( 11)),((i,j,k)) ) (x^5 )^i (x^2 )^j (1)^k = Σ_(i+j+k ≤ 11) ^(11) ((( 11)),(( i,j,k)) ) x^(5i+2j) (1)^k i = 0 ⇒j = 0 ∧k =11, j = 1 ∧k = 9 , j=2 ∧ k = 7 j = 3 ∧ k = 5 , j = 4 ∧k= 3 j = 5 ∧ k = 1 i=1 ⇒ j = 0 ∧k = 6 , j = 1 ∧k= 4 j = 2 ∧ k = 2 , j = 3 ∧ k = 0 i = 2 ⇒ j = 0∧k=1 Σ terms = 11

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

Commented by jagoll last updated on 07/Apr/20

hehehe

$$\mathrm{hehehe} \\ $$

Commented by ajfour last updated on 07/Apr/20

I still think, mrW Sirs, obvious answer, correct.

$${I}\:{still}\:{think},\:{mrW}\:{Sirs},\:{obvious} \\ $$$${answer},\:{correct}. \\ $$

Commented by jagoll last updated on 08/Apr/20

what answer this question sir?

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

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

when expanded, (x^5 +x^2 +1)^(11) has 51 terms. (x^5 +x^2 +1)^(11) =Σ_(i+j+k=11) (_(i,j,k) ^(11) )(x^5 )^i (x^2 )^j 1^k =Σ_(i+j+k=11) (_(i,j,k) ^(11) )x^(5i+2j) i,j,k≥0 i+j+k=11 ⇒0≤i+j≤11 0≤5i+2j≤55 but from 0 to 55 there are only 51 numbers which is a multiple of 5 (max. 55) or a multiple of 2 (max. 22) or the sum of both. following terms don′t exist: x^1 , x^3 , x^(48) , x^(51) , x^(53) because you can′t express the numbers 1, 3, 48, 51, 53 as 5i+2j with 0≤i+j≤11.

$${when}\:{expanded},\:\left({x}^{\mathrm{5}} +{x}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{11}} \:{has}\:\mathrm{51}\: \\ $$$${terms}. \\ $$$$ \\ $$$$\left({x}^{\mathrm{5}} +{x}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{11}} =\underset{{i}+{j}+{k}=\mathrm{11}} {\sum}\left(_{{i},{j},{k}} ^{\mathrm{11}} \right)\left({x}^{\mathrm{5}} \right)^{{i}} \left({x}^{\mathrm{2}} \right)^{{j}} \mathrm{1}^{{k}} \\ $$$$=\underset{{i}+{j}+{k}=\mathrm{11}} {\sum}\left(_{{i},{j},{k}} ^{\mathrm{11}} \right){x}^{\mathrm{5}{i}+\mathrm{2}{j}} \\ $$$${i},{j},{k}\geqslant\mathrm{0} \\ $$$${i}+{j}+{k}=\mathrm{11}\:\Rightarrow\mathrm{0}\leqslant{i}+{j}\leqslant\mathrm{11} \\ $$$$\mathrm{0}\leqslant\mathrm{5}{i}+\mathrm{2}{j}\leqslant\mathrm{55} \\ $$$${but}\:{from}\:\mathrm{0}\:{to}\:\mathrm{55}\:{there}\:{are}\:{only}\:\mathrm{51}\: \\ $$$${numbers}\:{which}\:{is}\:{a}\:{multiple}\:{of}\:\mathrm{5} \\ $$$$\left({max}.\:\mathrm{55}\right)\:{or}\:{a}\:{multiple}\:{of}\:\mathrm{2}\:\left({max}.\:\mathrm{22}\right) \\ $$$${or}\:{the}\:{sum}\:{of}\:{both}. \\ $$$$ \\ $$$${following}\:{terms}\:{don}'{t}\:{exist}: \\ $$$${x}^{\mathrm{1}} ,\:{x}^{\mathrm{3}} ,\:{x}^{\mathrm{48}} ,\:{x}^{\mathrm{51}} ,\:{x}^{\mathrm{53}} \\ $$$${because}\:{you}\:{can}'{t}\:{express}\:{the}\:{numbers} \\ $$$$\mathrm{1},\:\mathrm{3},\:\mathrm{48},\:\mathrm{51},\:\mathrm{53}\:{as}\:\mathrm{5}{i}+\mathrm{2}{j}\:{with}\:\mathrm{0}\leqslant{i}+{j}\leqslant\mathrm{11}. \\ $$

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