Let f(x) = 2x^n + lambda,where lambda in R an | vedclass

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(B) Given the function $f(x) = 2x^n + \lambda$. Using the given values: $f(4) = 2(4^n) + \lambda = 133$  --- $(1)$ $f(5) = 2(5^n) + \lambda = 255$  --

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$60$

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(B) Given the function $f(x) = 2x^n + \lambda$. Using the given values: $f(4) = 2(4^n) + \lambda = 133$  --- $(1)$ $f(5) = 2(5^n) + \lambda = 255$  --- $(2)$ Subtracting equation $(1)$ from $(2)$: $2(5^n - 4^n) = 255 - 133 = 122$ $5^n - 4^n = 61$ By testing values for $n \in N$: For $n = 3$,$5^3 - 4^3 = 125 - 64 = 61$. Thus,$n = 3$. Substituting $n = 3$ into equation $(1)$: $2(4^3) + \lambda = 133$ $2(64) + \lambda = 133$ $128 + \lambda = 133 \Rightarrow \lambda = 5$. Now,calculate $f(3) - f(2)$: $f(3) = 2(3^3) + 5 = 2(27) + 5 = 59$ $f(2) = 2(2^3) + 5 = 2(8) + 5 = 21$ $f(3) - f(2) = 59 - 21 = 38$. The divisors of $38$ are $1, 2, 19, 38$. The sum of these divisors is $1 + 2 + 19 + 38 = 60$.

Let $f(x) = 2x^n + \lambda$,where $\lambda \in R$ and $n \in N$. Given $f(4) = 133$ and $f(5) = 255$,find the sum of all the positive integer divisors of $(f(3) - f(2))$.

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