The number of dissimilar terms in the expansion of (x + y)n is n + 1. Therefore number of dissimilar terms in the expansion of (x + y + z)12 is

Question

The number of dissimilar terms in the expansion of (x + y)n is n + 1. Therefore number of dissimilar terms in the expansion of (x + y + z)12 is (A) 13 (B) 39 (C) 78 (D) 91

Tardigrade Admin · Accepted Answer

(x + y + z)12 = x12 + 12c1x11(y + z)1 + 12c2x10 (y + z)2 + ..... + 12c12(y + z)12 ∴ total no. of terms = 1+2 + 3 + 4 +..... + 13 = (13/2)(1+13) = 13 × 7 = 91

Q. The number of dissimilar terms in the expansion of $(x + y)^{n}$ is $n + 1$ . Therefore number of dissimilar terms in the expansion of $(x + y + z)^{12}$ is

13

39

78

91

$(x + y + z)^{12} = x^{12} +^{12} c_{1} x^{11} (y + z)^{1}$
$+^{12} c_{2} x^{10} (y + z)^{2} + ..... +^{12} c_{12} (y + z)^{12}$
$∴$ total no. of terms $= 1 + 2 + 3 + 4 + ..... + 13$
$= \frac{13}{2} (1 + 13) = 13 \times 7 = 91$

Q. The number of dissimilar terms in the expansion of (x+y)n is n+1. Therefore number of dissimilar terms in the expansion of (x+y+z)12 is

13

39

78

91

(x+y+z)12=x12+12c1​x11(y+z)1 +12c2​x10(y+z)2+.....+12c12​(y+z)12 ∴ total no. of terms =1+2+3+4+.....+13 =213​(1+13)=13×7=91

Q. The number of dissimilar terms in the expansion of $(x + y)^{n}$ is $n + 1$ . Therefore number of dissimilar terms in the expansion of $(x + y + z)^{12}$ is

$(x + y + z)^{12} = x^{12} +^{12} c_{1} x^{11} (y + z)^{1}$
$+^{12} c_{2} x^{10} (y + z)^{2} + ..... +^{12} c_{12} (y + z)^{12}$
$∴$ total no. of terms $= 1 + 2 + 3 + 4 + ..... + 13$
$= \frac{13}{2} (1 + 13) = 13 \times 7 = 91$