What is the simplified value of (3+1)(3^{2}+1 | vedclass

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Question

(A) To simplify the expression $(3+1)(3^{2}+1)(3^{4}+1)(3^{8}+1)(3^{16}+1)$,we use the algebraic identity $(a-b)(a+b) = a^{2}-b^{2}$.Multiply and divi

Vedclass · Accepted Answer

$\frac{3^{32}-1}{2}$

Vedclass · Answer

(A) To simplify the expression $(3+1)(3^{2}+1)(3^{4}+1)(3^{8}+1)(3^{16}+1)$,we use the algebraic identity $(a-b)(a+b) = a^{2}-b^{2}$.Multiply and divide the expression by $(3-1)$,which is $2$:$= \frac{(3-1)(3+1)(3^{2}+1)(3^{4}+1)(3^{8}+1)(3^{16}+1)}{3-1}$$= \frac{(3^{2}-1)(3^{2}+1)(3^{4}+1)(3^{8}+1)(3^{16}+1)}{2}$$= \frac{(3^{4}-1)(3^{4}+1)(3^{8}+1)(3^{16}+1)}{2}$$= \frac{(3^{8}-1)(3^{8}+1)(3^{16}+1)}{2}$$= \frac{(3^{16}-1)(3^{16}+1)}{2}$$= \frac{3^{32}-1}{2}$.

What is the simplified value of $(3+1)(3^{2}+1)(3^{4}+1)(3^{8}+1)(3^{16}+1)?$

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