Let $n \in N$ , such that $(1 + x + x^{2})^{n} = a_{0} + a_{1} x + a_{2} x^{2} . . . a_{2 n} x^{2 n}$ The value of $a_{0} + a_{1} + a_{2} . . . . a_{n - 1}$ is

\frac{1}{2} (3^{n})

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

\frac{1}{2} (3^{n} - a_{n})

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

\frac{a_{n}}{2}

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

3. {a_{n}}^{n}

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

Open in App

Solution

The correct option is B $\frac{1}{2} (3^{n} - a_{n})$
${(1 + x + x^{2})}^{n}$ = $\sum_{r = 0}^{n} a_{r} . x^{r}$ ...........eqn 1
We replace x by $\frac{1}{x}$
${(1 + \frac{1}{x} + \frac{1}{x^{2}})}^{n}$ = $\sum_{r = 0}^{n} a_{r} {(\frac{1}{x})}^{r}$
${(1 + x + x^{2})}^{n}$ = $\sum_{r = 0}^{n} a_{r} {(x)}^{2 n - r}$
$\sum_{r = 0}^{n} a_{r} . x^{r}$ = $\sum_{r = 0}^{n} a_{r} {(x)}^{2 n - r}$ ............ using eqn 1
Equating the coefficent of $x^{2 n - r}$ on both sides we get
So $a_{r} = a_{2 n - r}$
Now if we put x=1 we get
${(1 + 1 + 1)}^{n}$ = $a_{0} + a_{1} + a_{2} . . . . . . . . . . . . . . . . . . . + a_{2 n}$
$3^{n}$ = $a_{0} + a_{1} + a_{2} . . . . . . . . . . . . . . . . . . . + a_{2 n}$
But $a_{r} = a_{2 n - r}$
$3^{n}$ = $2 (a_{0} + a_{1} + a_{2} . . . . . . . . . . . . . . . . . . . + a_{n - 1}) + a_{n}$
$3^{n} - a_{n}$ = $2 (a_{0} + a_{1} + a_{2} . . . . . . . . . . . . . . . . . . . + a_{n - 1})$
$\frac{3^{n} - a_{n}}{2}$ = $(a_{0} + a_{1} + a_{2} . . . . . . . . . . . . . . . . . . . + a_{n - 1})$

Suggest Corrections

Similar questions

{(1 + x + x^{2})}^{n} = a_{0} + a_{1} x + a_{2} x^{2} + \dots + a_{2 n} x^{2 n},

a_{0} + a_{1} + a_{2} + \dots . + a_{2 n}

a_{0} - a_{1} + a_{2} - a_{3} \dots . . + a_{2 n}

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

a_{0}, a_{1}, a_{2}, . . . . . . . . . . ., a_{2 n}

a_{n}

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

a_{0}, a_{1}, a_{2}, . . ., a_{2 n}

a_{n} = \frac{1}{2 n + 1}

n \in N

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

a_{r}

(0 \leq r \leq 2 n)

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

a_{0} + a_{2} + a_{4} + . . . . + a_{2 n}

Join BYJU'S Learning Program