Answer to Question #157288 in Calculus for Vishal

Question #157288

Use induction to prove that 1³+2³+...+(n−1)³ < n⁴/4 < 1³+2³+...+n³


1
Expert's answer
2021-02-28T07:14:19-0500

1³+2³+...+(n−1)³ < n⁴/4 < 1³+2³+...+n³

For n=1,

"0 < \\frac{1}{4}<1\\\\" This is true

We assume it is true for n=k, then

1³+2³+...+(k−1)³ < k⁴/4 < 1³+2³+...+k³

For n = k+1,

We show that

1³+2³+...+(k)³ < (k+1)⁴/4 < 1³+2³+...+k³+(k+1)³

Add k³ to (1)

1³+2³+...+(k−1)³ < k⁴/4 + k³ <1³+2³+...+k³+k³

But k⁴/4 + k³< "\\frac{k^4+4k^3+6k^2+4k+1}{4}\\\\"

k⁴/4 + k³< k⁴/4 + k³+ "\\frac{6k^2+4k+1}{4}=\\frac{(k+1)^4}{4}\\\\\n\\implies"

1³+2³+...+(k)³ < k⁴/4 + k³< (k+1)⁴/4 < 1³+2³+...+k³+(k+1)³,

and since k³ < (k+1)³,

1³+2³+...+(k)³ < (k+1)⁴/4 < 1³+2³+...+k³+(k+1)³

Which is true for n=k+1


Hence the statement is true.


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