Question

Find the coefficient of
${x}^{21} \: \: in \: {(1 + x + {x}^{2} +..... + {x}^{9}) }^{ 3}$

Answer 1

$its \: 1 \\ the \: coefficient \: of \: {x}^{21} \\ it \: can \: be \: written \: as \: \: {x}^{7} \times {x}^{3} \\ \\ hope \: this \: helps \: you$

Answer 2

$coefficient \: of \: {x}^{21} \\ \\ 1 + x + {x}^{2} ..... + {x}^{9} \: is \: a \: gp \\ hence \: it \: can \: be \: simplified \: as \: \\ \frac{(1 - {x}^{10}) }{1 - x} \\ so \: its \: cube = {(1 - {x}^{10}) }^{3} {(1 - x)}^{ - 3} \\ we \: have \: to \: find \: the \: coefficint \: of \\ \: {x}^{21} \\ firstly \: simplify \: it \: more \\ (1 + 3 {x}^{20} - 3 {x}^{10} ) {(1 - x)}^{ - 3} \\ hence \\ \binom{3 + 21 - 1}{21} + 3 \binom{3}{1} - 3 \binom{13}{11} \\ \frac{23 \times 22}{2} + 9 - 3 \times \frac{13 \times 12}{2} \\ 262 - 234 = 28$
hence it's coefficient will be 28