TBQ - Evaluate the sum of the numbers — task. Maths TNSB Mentoring, Class 10 Crash course.

If \(1^3 + 2^3 + … + k^3\) \(=\) \(189225\), then find \( 1+ 2 + 3 + … + k\).

Answer:

Sum of the cubes of first \(n\) natural numbers \(=\)

{(\frac{i (i + i)}{i})}^{i}

\( 1+ 2 + 3 + … + k\) \(=\)

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5. TBQ - Evaluate the sum of the numbers