Question

50 points

No. of terms in $\sf\:(1+3x+3x^2 + x^3)^6$ is

don't spam​

Answer 1

Answer:

No. of terms in $\sf\:(1+3x+3x^2 + x^3)^6$ is 19

Step-by-step explanation:

$\sf\:(1+3x+3x^2 + x^3)^6$

=$\sf\:[1^{3}+3\times1^{2}\times x +3\times 1\times x^{2}+x^{3}]^{6}$

=$\sf\:[(1+x)^{3}]^{6}$

/* we know the algebraic identity:

a³+3a²b+3ab²+b²=(a+b)³ */

=$\sf\:(1+x)^{3\times 6}$

=$\sf\:(1+x)^{18}$

No. of terms in $\sf\:(1+3x+3x^2 + x^3)^6$ = 18+1 = 19 terms

/* Number of terms in the expansion of (1+x)ⁿ = n+1 */

Therefore,

No. of terms in $\sf\:(1+3x+3x^2 + x^3)^6$ is 19

•••♪

Answer 2

Answer:

19

Step-by-step explanation:

Hope it helps you!..................