Algebra
Binomial Theorem / Coefficients
MJMT_Full_Test_10
Grade 12
Question:
Let $(1+x+x^2)^{30} = \displaystyle\sum_{r=0}^{60} a_r x^r$. If $\alpha a_{21} = \beta a_{20} + \gamma a_{19}$, $(\alpha,\beta,\gamma\in\mathbb{N})$, then $\alpha+\beta+\gamma$ can be
Step-by-Step Solution
Key Concept: Use the recurrence relation for coefficients of $(1+x+x^2)^n$: $(r+1)a_{r+1}=(2n-r+1)a_r+\ldots$; differentiate the generating function.
Differentiating $(1+x+x^2)^{30}$: $(1+2x)(1+x+x^2)^{30}=\sum ra_r x^{r-1}\cdot(1+x+x^2)$... Using the recurrence $21a_{21}=41a_{20}+10a_{19}$ gives $\alpha+\beta+\gamma=72$.
Correct Answer: 2