Answer to Question #157288 in Calculus for Vishal

Question

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

Answer 1

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+ $\frac{6k^2+4k+1}{4}=\frac{(k+1)^4}{4}\\ \implies$

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.