Which of the following statements is/are correct?

(a) The sum $$1^3+2^3+3^3+$$​ ... to ten terms, lies between 3020 and 3030.

(b) If $$x$$​ is subtracted from the sum $$1^3+3^3+5^3+$$​ ... to 10 terms, then it becomes a square. The least value of $$x$$​ is 17.

(c) $$11^3+12^3+13^3+$$​ ... $$+20^3=41075$$​.

Correct option is C

Solution:

Statement (a):

Sum of first 10 cubes:
1³ + 2³ + 3³ + ... + 10³

Formula:

Sum of cubes of first n natural numbers :

​$$\left[\frac{n(n + 1)}{2}\right]^2 \\\$$​​

Using n = 10:

​$$\\\left[\frac{10 \times 11}{2}\right]^2 = (55)^2 = 3025$$​​

3025 lies between 3020 and 3030
Statement (a) is correct

Statement (b):

Sum of first 10 odd cubes:
1³ + 3³ + 5³ + ... + 19³

Compute actual sum:
= 1 + 27 + 125 + 343 + 729 + 1331 + 2197 + 3375 + 4913 + 6859 = 19800

Check if (19800 − x) is a perfect square with x = 17:

​$$\\19800 - 17 = 19783 \\\sqrt{19783} \approx 140.6 \notin \mathbb{Z}$$​​

which is not a whole number
Statement (b) is incorrect

Statement (c):

Sum from 11³ to 20³
Use formula:
​$$\\\text{Sum from } 11^3 \text{ to } 20^3 = \left[\frac{20 \cdot 21}{2}\right]^2 - \left[\frac{10 \cdot 11}{2}\right]^2 \\\ \\= (210)^2 - (55)^2 = 44100 - 3025 = 41075 \\$$​​

So, required sum = 44100 − 3025 = 41075
Statement (c) is correct

Correct Statements: (a) and (c)

​Final Answer: (c)

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