If $(1 + 2 x + 3 x^{2})^{10} = a_{0} + a_{1} x + a_{2} x^{2} + . . . + a_{20} x^{20}$ , then $a_{1}$ equals

10

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

20

No worries! We‘ve got your back. Try BYJU‘S free classes today!

210

No worries! We‘ve got your back. Try BYJU‘S free classes today!

None of these

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $10$
Given,
$(1 + 2 x + 3 x^{2})^{10} = a_{0} + a_{1} x + a_{2} x^{2} + . . . + a_{20} x^{20}$

Then, $a_{1}$ = Coefficient of $x$ in $(1 + 2 x + 3 x^{2})^{10}$

Now,
$(1 + 2 x + 3 x^{2})^{10} = {(1 + 2 x) + 3 x^{2}}^{10}$
$=^{10} C_{0} (1 + 2 x)^{10} +^{10} C_{1} (1 + 2 x)^{9} (3 x^{2}) + . . .$ .

$∴$ Coefficient of $x$ in $(1 + 2 x + 3 x^{2})^{10}$
$=$ Coefficient of $x$ in $^{10} C_{0} (1 + 2 x)^{10}$
$=^{10} C_{0} \cdot 2 \cdot^{10} C_{1}$
$= 1 \cdot 2 \cdot 10 = 20$

Suggest Corrections

Similar questions

(1 + 2 x + 3 x^{2})^{10} = a_{0} + a_{1} x + a_{2} x^{2} + . . . + a_{20} x^{20},

a_{1}

{(1 + 2 x + 3 x^{2})}^{10} = a_{0} + a_{1} x + a_{2} x^{2} + . . . . . + a_{20} x^{20}

a_{1}

(1 + 2 x)^{20} = a_{0} + a_{1} x + a_{2} x^{2} + . . . . + a_{20} x^{20}

3 a_{0} + 2 a_{1} + 3 a_{2} + 2 a_{3} + 3 a_{4} + 2 a_{5} + . . . 2 a_{19} + 3 a_{20}

(1 + 2 x)^{20} = a_{0} + a_{1} x + a_{2} x^{2} + . + a_{20} x^{20}

3 a_{0} + 2 a_{1} + 3 a_{2} + 2 a_{3}, + 3 a_{4} + 2 a_{2} + . . . . 2 a_{19} + 3 a_{20}

i f (1 + x)^{10} = a_{0} + a_{1} x + . . . . a_{10} x^{10}, t h e n (a_{0} - a_{2} + a_{4} - a_{6} + a_{8} - a_{10})^{2} + (a_{1} - a_{3} + a_{5} - a_{7} + a_{9})^{2}

Join BYJU'S Learning Program