Let the coefficient of \(\mathrm{x}^{\mathrm{r}}\) in the expansion of \[ \begin{aligned} & (\mathrm{x}+3)^{\mathrm{n}-1}+(\mathrm{x}+3)^{\mathrm{n}-2}(\mathrm{x}+2)+ \\ & (\mathrm{x}+3)^{\mathrm{n}-3}(\mathrm{x}+2)^{2}+\ldots \ldots+(\mathrm{x}+2)^{\mathrm{n}-1} \end{aligned} \] be \(\alpha_{r}\). If \(\sum_{r=0}^{n} \alpha_{r}=\beta^{n}-\gamma^{n}, \beta, \gamma \in N\), then the value of \(\beta^{2}+\gamma^{2}\) equals ______.

JEE Main · 2024 · 31 Jan 2024 · Shift 2 · Binomial Theorem

Let the coefficient of

x^{r}

in the expansion of

(x + 3)^{n - 1} + (x + 3)^{n - 2} (x + 2) + (x + 3)^{n - 3} (x + 2)^{2} + \dots\dots + (x + 2)^{n - 1}

α_{r}

. If

\sum_{r = 0}^{n} α_{r} = β^{n} - γ^{n}, β, γ \in N

, then the value of

β^{2} + γ^{2}

equals ______.

Answer: 25

The given series is a geometric progression in the ratio

\frac{x + 2}{x + 3}

, with

n

terms, first term

(x + 3)^{n - 1}

and last term

(x + 2)^{n - 1}

. Summing it as a GP with common ratio

r = \frac{x + 2}{x + 3}

k = 0 \sum n - 1 (x + 3)^{n - 1 - k} (x + 2)^{k} = \frac{( x + 3 ) ^{n} - ( x + 2 ) ^{n}}{( x + 3 ) - ( x + 2 )} = (x + 3)^{n} - (x + 2)^{n}

.

So

r \sum α_{r} x^{r} = (x + 3)^{n} - (x + 2)^{n}

.

Putting

x = 1

gives the sum of all the coefficients:

r = 0 \sum n α_{r} = 4^{n} - 3^{n}

.

Comparing with

β^{n} - γ^{n}

(

β, γ \in N

) gives

β = 4, γ = 3

β^{2} + γ^{2} = 16 + 9 = 25

JEE Main · 2024 · 31 Jan 2024 · Shift 2 · Integer answer