In the expansion of(1+x+x2+x3)6, the coefficient of x14 is
130
120
128
125
115
Explanation for the correct option:
Finding the coefficient of x14:
(1+x)+x2(1+x)6Rewritingtheequation,weget⇒(1+x)(1+x2)6⇒(1+x)6(1+x2)6
Expanding it with the help of binomial expansion
We know that x+yn=C0nxny0+C1nxn-1y1+C2nxn-2y2+……………+Cnnx0yn
=6C0+6C1x+6C2x2+6C3x3+6C4x4+6C5x5+6C6x66C0+6C1x2+6C2x4+6C3x6+6C4x8+6C5x10+6C6x12=6C26C6x2+12+6C46C5x4+10+6C66C4x6+8+.......consideringonlythetermswithx14=6C26C6+6C46C5+6C66C4x14+.......=6!2!4!6!6!0!+6!4!2!6!5!1!+6!6!0!6!4!2!x14+.......=6×5×4×3×2×11×2×4×3×2×11+6×5×4×3×2×14×3×2×1×2×16×5×4×3×2×15×4×3×2×1×1+16×5×4×3×2×14×3×2×1×2×1=15+15(6)+15x14=120x14
Hence, option B is correct.
The coefficient of the term independent of x in the expansion of (ax+bx)14 is