In VVi+1-2 dt, if we let V Vt+1 – 2, then dt is equal to A 2(x²+2)dx ® 4x(x²+2)dx c) None D (4x³+2)dx E 4x³dx

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A: solution is given below

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A: We have to find out

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A: an=n45nlimn→∞ an+1an=limn→∞ (n+1)4 5n5n+1 n4=limn→∞ (n+1)4 5nn4  5n 5

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