Question:

Step 1: Recall the formula for the sum of the cubes of the first $n$ natural numbers. The sum of the cubes of the first $n$ natural numbers is given by the formula: $$ \sum_{k=1}^{n} k^3 = \left[ \frac{n(n+1)}{2} \right]^2 $$ Step 2: Calculate the sum of the cubes of the first 50 natural numbers. Using the formula from Step 1 with $n=50$, we calculate $\sum_{k=1}^{50} k^3$: $$ \sum_{k=1}^{50} k^3 = \left[ \frac{50(50+1)}{2} \right]^2 = \left[ \frac{50 \times 51}{2} \right]^2 = \left[ 25 \times 51 \right]^2 = [1275]^2 = 1,625,625 $$ Step 3: Calculate the sum of the cubes of the first 30 natural numbers. The given series starts from $31^3$. To find the sum from $31^3$ to $50^3$, we need to subtract the sum of the cubes from $1^3$ to $30^3$ from the sum of the cubes from $1^3$ to $50^3$. Using the formula from Step 1 with $n=30$, we calculate $\sum_{k=1}^{30} k^3$: $$ \sum_{k=1}^{30} k^3 = \left[ \frac{30(30+1)}{2} \right]^2 = \left[ \frac{30 \times 31}{2} \right]^2 = \left[ 15 \times 31 \right]^2 = [465]^2 = 216,225 $$ Step 4: Calculate the sum of the series $31^3 + 32^3 + \cdots + 50^3$. The desired sum can be found by subtracting the sum of the first 30 cubes (calculated in Step 3) from the sum of the first 50 cubes (calculated in Step 2): $$ \sum_{k=31}^{50} k^3 = \sum_{k=1}^{50} k^3 - \sum_{k=1}^{30} k^3 $$ $$ \sum_{k=31}^{50} k^3 = 1,625,625 - 216,225 = 1,409,400 $$ The sum of the series $31^3 + 32^3 + \cdots + 50^3$ is $1,409,400$.