Question

Let $(1 + x + x^2)^{30} = \sum_{r=0}^{60} a_r x^r$. If $\alpha a_{21} = \beta a_{20} + \gamma a_{19}, (\alpha, \beta, \gamma \in N)$ then the value of $\alpha + \beta + \gamma$ can be

• A. 62
• B. 72
• C. 82
• D. 74

Answer 1

Answer: (B)

72

Answer 2

The core of the solution involves deriving a recurrence relation between the coefficients of the polynomial expansion $(1 + x + x^2)^{30} = \sum_{r=0}^{60} a_r x^r$. This is achieved by differentiating the polynomial and manipulating the resulting equation.

1. Define $P(x) = (1 + x + x^2)^{30}$.
2. Differentiate $P(x)$ to get $P'(x) = 30(1 + x + x^2)^{29}(1 + 2x)$.
3. Establish the relation $P'(x)(1+x+x^2) = 30(1+2x)P(x)$.
4. Substitute the series expansions for $P(x)$ and $P'(x)$ into this relation: $(\sum r a_r x^{r-1})(1+x+x^2) = 30(1+2x)(\sum a_r x^r)$.
5. Equate the coefficients of $x^n$ on both sides. The general relation obtained is $(n+1)a_{n+1} = (30-n)a_n + (61-n)a_{n-1}$.
6. To match the given form $\alpha a_{21} = \beta a_{20} + \gamma a_{19}$, we set $n+1=21$ (so $n=20$).
7. Substituting $n=20$ yields $21 a_{21} = 10 a_{20} + 41 a_{19}$.
8. By comparison, $\alpha=21, \beta=10, \gamma=41$. These are natural numbers.
9. Calculate $\alpha + \beta + \gamma = 21 + 10 + 41 = 72$.