If the coefficients of x^{10} and x^{11} in t | vedclass

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(B) The expansion is $(1+\alpha x+\beta x^2)(1+x)^{11}$.Using the binomial expansion $(1+x)^{11} = \sum_{k=0}^{11} \binom{11}{k} x^k$,the expression b

Vedclass · Accepted Answer

$13$

Vedclass · Answer

(B) The expansion is $(1+\alpha x+\beta x^2)(1+x)^{11}$.Using the binomial expansion $(1+x)^{11} = \sum_{k=0}^{11} \binom{11}{k} x^k$,the expression becomes:$(1+\alpha x+\beta x^2) \sum_{k=0}^{11} \binom{11}{k} x^k = \sum \binom{11}{k} x^k + \alpha \sum \binom{11}{k} x^{k+1} + \beta \sum \binom{11}{k} x^{k+2}$.For $x^{10}$,the coefficient is $\binom{11}{10} + \alpha \binom{11}{9} + \beta \binom{11}{8} = 11 + 55\alpha + 165\beta = 396$.Dividing by $11$,we get $1 + 5\alpha + 15\beta = 36$,so $5\alpha + 15\beta = 35$,or $\alpha + 3\beta = 7$ (Equation $1$).For $x^{11}$,the coefficient is $\binom{11}{11} + \alpha \binom{11}{10} + \beta \binom{11}{9} = 1 + 11\alpha + 55\beta = 144$.So,$11\alpha + 55\beta = 143$,or $\alpha + 5\beta = 13$ (Equation $2$).Subtracting Equation $1$ from Equation $2$: $(5\beta - 3\beta) = 13 - 7$,so $2\beta = 6$,which means $\beta = 3$.Substituting $\beta = 3$ into Equation $1$: $\alpha + 3(3) = 7$,so $\alpha = 7 - 9 = -2$.Thus,$\alpha^2 + \beta^2 = (-2)^2 + 3^2 = 4 + 9 = 13$.

If the coefficients of $x^{10}$ and $x^{11}$ in the expansion of $(1+\alpha x+\beta x^2)(1+x)^{11}$ are $396$ and $144$ respectively,then $\alpha^2+\beta^2=$

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