The value of 4 {^nC_1 + 4 cdot ^nC_2 + 4^2 cd | vedclass

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(D) Let $E = 4 \{^nC_1 + 4 \cdot ^nC_2 + 4^2 \cdot ^nC_3 + \dots + 4^{n-1} \cdot ^nC_n\}$.Distributing the $4$ inside the braces,we get:$E = 4 \cdot ^

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$5^n - 1$

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(D) Let $E = 4 \{^nC_1 + 4 \cdot ^nC_2 + 4^2 \cdot ^nC_3 + \dots + 4^{n-1} \cdot ^nC_n\}$.Distributing the $4$ inside the braces,we get:$E = 4 \cdot ^nC_1 + 4^2 \cdot ^nC_2 + 4^3 \cdot ^nC_3 + \dots + 4^n \cdot ^nC_n$.Recall the Binomial Theorem: $(1 + x)^n = ^nC_0 + ^nC_1 x + ^nC_2 x^2 + \dots + ^nC_n x^n$.Setting $x = 4$,we have:$(1 + 4)^n = ^nC_0 + ^nC_1(4) + ^nC_2(4^2) + \dots + ^nC_n(4^n)$.$5^n = 1 + (4 \cdot ^nC_1 + 4^2 \cdot ^nC_2 + \dots + 4^n \cdot ^nC_n)$.$5^n = 1 + E$.Therefore,$E = 5^n - 1$.

The value of $4 \{^nC_1 + 4 \cdot ^nC_2 + 4^2 \cdot ^nC_3 + \dots + 4^{n-1} \cdot ^nC_n\}$ is:

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