Exams Joint Entrance Examination Mathematics Binomial Theorem Question 79 Question 79 of 347 Medium NUMERICAL
Let the coefficient of x r x^r x r in the expansion of \[(x+3)^{n-1}+(x+3)^{n-2}(x+2)+(x+3)^{n-3}(x+2)^2+\ldots \ldots \ldots .+(x+2)^{n-1}\] be \[\alpha_r\]. If \[\sum_\limits{r=0}^n \alpha_r=\beta^n-\gamma^n, \beta, \gamma \in \mathbb{N}\], t...
Answer: 25
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( x + 3 ) n − 1 + ( x + 3 ) n − 2 ( x + 2 ) + ( x + 3 ) n − 3 ( x + 2 ) 2 + … … . + ( x + 2 ) n − 1 ∑ α r = 4 n − 1 + 4 n − 2 × 3 + 4 n − 3 × 3 2 … … + 3 n − 1 = 4 n − 1 [ 1 + 3 4 + ( 3 4 ) 2 … . + ( 3 4 ) n − 1 ] = 4 n − 1 × 1 − ( 3 4 ) n 1 − 3 4 = 4 n − 3 n = β n − γ n β = 4 , γ = 3 β 2 + γ 2 = 16 + 9 = 25 \begin{gathered} (x+3{)}^{n-1}+(x+3{)}^{n-2}(x+2)+(x+3{)}^{n-3} \\ (x+2{)}^{2}+\ldots \ldots .+(x+2{)}^{n-1} \\ \sum {\alpha }_{r}={4}^{n-1}+{4}^{n-2}\times 3+{4}^{n-3}\times {3}^{2}\ldots \ldots +{3}^{n-1} \\ ={4}^{n-1}[1+\frac{3}{4}+{(\frac{3}{4})}^{2}\ldots .+{(\frac{3}{4})}^{n-1}] \\ ={4}^{n-1}\times \frac{1-{(\frac{3}{4})}^{n}}{1-\frac{3}{4}} \\ ={4}^{n}-{3}^{n}={\beta }^{n}-{\gamma }^{n} \\ \beta =4,\gamma =3 \\ {\beta }^{2}+{\gamma }^{2}=16+9=25 \end{gathered} ( x + 3 ) n − 1 + ( x + 3 ) n − 2 ( x + 2 ) + ( x + 3 ) n − 3 ( x + 2 ) 2 + …… . + ( x + 2 ) n − 1 ∑ α r = 4 n − 1 + 4 n − 2 × 3 + 4 n − 3 × 3 2 …… + 3 n − 1 = 4 n − 1 [ 1 + 4 3 + ( 4 3 ) 2 … . + ( 4 3 ) n − 1 ] = 4 n − 1 × 1 − 4 3 1 − ( 4 3 ) n = 4 n − 3 n = β n − γ n β = 4 , γ = 3 β 2 + γ 2 = 16 + 9 = 25 Answer the question to unlock