You visited us 1 times! Enjoying our articles? Unlock Full Access!

Sum of Coefficients of All Terms

If a0 , a...

Question

If $a_{0}, a_{1}, a_{2}$ ....be the coefficients in the expansion of $(1 + x + x$ $^{2})^{n}$ in ascending powers $f (x)$ , then prove that :
$a_{r} = a_{2 n - r}$

Open in App

Solution

$(1 + x + x^{2})^{n} = a_{0} + a_{1} x + a_{2} x^{2} + \dots + a_{2 n} x^{2 n} = \sum_{r = 0}^{2 n} a_{r} x^{r}$
Replace $x$ by $\frac{1}{x}$ .
${(1 + \frac{1}{x} + \frac{1}{x^{2}})}^{2} = a_{0} + a_{1} (\frac{1}{x}) + a_{2} (\frac{1}{x^{2}}) + \dots$
or $\frac{(1 + x + x^{2})^{n}}{x^{2 n}} = a_{0} + a_{1} (\frac{1}{x}) + \dots + a_{r} (\frac{1}{x^{r}}) + \dots$
or $(1 + x + x^{2})^{n} = a_{0} x^{2 n} + a_{1} x^{2 n - 1} + \dots + a_{r} x^{2 n - r} + \dots$
$\sum_{r = 0}^{2 n} a_{r} x^{r} = \sum_{r = 0}^{2 n} a_{r}^{2 n + r}$ , by (1)(2)
Equating the coefficients of $x^{2 n - r}$ in both sides, we get $a_{2 n - r} = a_{r} o r a_{r} = a_{2 n - r}$ , where 0 < r < 2n

Suggest Corrections

Similar questions

a_{0}, a_{1}, a_{2}

^{2})^{n}

a_{0} + a_{1} + a_{2} + . . . . + a_{n - 1} = \frac{1}{3} (3^{n} - a_{n})

a_{0}, a_{1}, a_{2}

(1 + x + x

^{2})^{n}

f (x)

(r + 1) a_{r + 1} = (n - r) a_{r} + (2 n - r + 1) a_{r - 1}, (0 < r < 2 n)

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

(1 + x + x^{2})^{n}

(a_{0} + a_{3} + a_{6} + . . .) = (a_{1} + a_{4} + a_{7} + . . . .) = = (a_{2} + a_{5} + a_{8} + . . .) = 3^{n - 1}

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

{(1 + x + x^{2})}^{n}

x

a_{0} a_{1} - a_{1} a_{2} + a_{2} a_{3} - . . . . =

a_{0}, a_{1}, a_{2}

^{2})^{n}

a_{0}^{2} - a_{1}^{2} + a_{2}^{2} - a_{3}^{2} + . . . . . . + (- 1)^{n - 1} a_{n - 1}^{2}

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

Join BYJU'S Learning Program