Let (1 + x + 2x2)20 = a0 + a1x + a2x2 + ....

Question

Let (1 + x + 2x2)20 = a0 + a1x + a2x2 + .... + a40x40. Then a1 + a3 + a5 + ..... + a37 is equal to

Accepted Answer

$${(1 + x + 2{x^2})^{20}} = {a_0} + {a_1}x + {a_2}{x^2} + .... + {a_{40}}{x^{40}}$$

Put x = 1

$$ \Rightarrow {4^{20}} = {a_0} + {a_1} + ....... + {a_{40}}$$ ..... (i)

Put x = -1

$$ \Rightarrow {2^{20}} = {a_0} - {a_1} + ....... +  - {a_{39}} + {a_{40}}$$ ..... (ii)

by (i) - (ii) we get,

$${4^{20}} - {2^{20}} = 2({a_1} + {a_3} + ...... + {a_{37}} + {a_{39}})$$

$$ \Rightarrow {a_1} + {a_3} + ...... + {a_{37}} = {2^{39}} - {2^{19}} - {a_{39}}$$ ..... (iii)

a39 = coeff. x39 in (1 + x + 2x2)20

$$ = {{20!} \over {0!1!9!}}{(1)^0}{(1)^1}{(2)^{19}}$$

$= {20.2^{19}}$

$$ \therefore $$ $${a_1} + {a_3} + ...... + {a_{37}} = {2^{39}} - {2^{19}}.21$$

$$ \Rightarrow {2^{19}}({2^{20}} - 21)$$

Solutions · 1