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The coefficient of $x^{8}$ in the expression ${(2 + x)}^{2} {(3 + x)}^{3} {(4 + x)}^{4}$ must be 29

The coefficient of $x^{203}$ in the expression $(x - 1) (x^{2} - 2) (x^{3} - 3) \dots .. (x^{20} - 20)$ must be 13

The coefficient of $x^{98}$ in the expression of $(x - 1) (x - 2) .. . (x - 100)$ must be $\frac{1}{2} [{(1 + 2 + 3 + ... + 100)}^{2} - (1^{2} + 2^{2} + 3^{2} + ... + 100^{2})]$

The coefficient of $x^{9}$ in the expansion of $(1 + x) (1 + x^{2}) (1 + x^{3}) \dots (1 + x^{100})$ is 9

answer is A, B, C.

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Detailed Solution

A. The expression ${(2 + x)}^{2} {(3 + x)}^{3} {(4 + x)}^{4} = (x + 2) (x + 2)$

$(x + 3) (x + 3) (x + 3) (x + 4) (x + 4) (x + 4) (x + 4) = x^{9} + (2 + 2 + 3 + 3 + 3 + 4 + 4 + 4 + 4) x^{8} + .....$

$\Rightarrow C o e f f i c i e n t o f x^{8} = 29$

B. Expression = $x . x^{2} . x^{3} ...... x^{20}$

$(1 - \frac{1}{x}) (1 - \frac{2}{x^{2}}) (1 - \frac{2}{x^{3}}) .... (1 - \frac{20}{x^{20}})$ Where $E = (1 - \frac{1}{x}) (1 - \frac{2}{x^{2}}) (1 - \frac{3}{x^{3}}) .... (1 - \frac{20}{x^{20}})$

Now coefficient of $x^{203}$ in original expression = coefficient of $x^{- 7}$ in E.
But, $E = 1 - (\frac{1}{x} + \frac{2}{x^{2}} + \frac{3}{x^{3}} .....) + (\frac{1}{x} . \frac{6}{x^{6}} + \frac{2}{x^{2}} . \frac{5}{x^{5}} + \frac{3}{x^{3}} . \frac{4}{x^{4}} + ....) - (\frac{1}{x} . \frac{2}{x^{2}} . \frac{4}{x^{4}} + ....)$

= coefficient of $x^{- 7} = - 7 + 6 + 10 + 12 - 8 = 13$

C. The coefficient of $x^{98} = (1.2 + 2.3 + \dots .) =$ Sum of product of first 100 natural numbers taken two at a time
$= \frac{1}{2} [{(1 + 2 + 3 + .... + 100)}^{2} - (1^{2} + 2^{2} + 3^{2} + ..... + 100^{2})]$

D . $x^{9}$ can be formed in 8 ways. i.e. $x^{9}, x^{1 + 8}, x^{2 + 7}, x^{3 + 6}, x^{4 + 5}, x^{1 + 2 + 6}, x^{1 + 3 + 5}, x^{2 + 3 + 4}$ and coefficient in
each case is 1.
$∴$ coefficient of $x^{9} = 1 + 1 + 1 + ....8$ times = 8