Let (1 + x + 2x2)20 = a0 + a1x + a2x2 + .... + a40x40. Then a1 + a3 + a5 + ..... + a37 is equal to
Solution
${(1 + x + 2{x^2})^{20}} = {a_0} + {a_1}x + {a_2}{x^2} + .... + {a_{40}}{x^{40}}$<br><br>Put x = 1<br><br>$\Rightarrow {4^{20}} = {a_0} + {a_1} + ....... + {a_{40}}$ ..... (i)<br><br>Put x = $-$1<br><br>$\Rightarrow {2^{20}} = {a_0} - {a_1} + ....... + - {a_{39}} + {a_{40}}$ ..... (ii)<br><br>by (i) $-$ (ii) we get,<br><br>${4^{20}} - {2^{20}} = 2({a_1} + {a_3} + ...... + {a_{37}} + {a_{39}})$<br><br>$\Rightarrow {a_1} + {a_3} + ...... + {a_{37}} = {2^{39}} - {2^{19}} - {a_{39}}$ ..... (iii)<br><br>a<sub>39</sub> = coeff. x<sup>39</sup> in (1 + x + 2x<sup>2</sup>)<sup>20</sup><br><br>$= {{20!} \over {0!1!9!}}{(1)^0}{(1)^1}{(2)^{19}}$<br><br>$= {20.2^{19}}$<br><br>$\therefore$ ${a_1} + {a_3} + ...... + {a_{37}} = {2^{39}} - {2^{19}}.21$<br><br>$\Rightarrow {2^{19}}({2^{20}} - 21)$
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Subject: Mathematics · Chapter: Binomial Theorem · Topic: Binomial Expansion
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