What is the sum of the series 13 – 23 + 33 – 43 + ........ + 93 eq

1.	What is the sum of the series 13 – 23 + 33 – 43 + ........ + 93 equal to?
Answer» Reqd. Sum = (1³ + 2³ + 3³ + 4³ + ..... + 9³) – 2(2³ + 4³ + 6³ + 8³) = (1³+ 2³ + 3³ + ..... + 9³) – 2⁴ (1³ + 2³ + 3³ + 4³) = \(\displaystyle\sum_{k=1}^{9} k^3\) - 2⁴ \(\displaystyle\sum_{k=1}^{4} k^3\) = \(\frac{9^2(9+1)^2}{4}\) - 2⁴ x \(\frac{4^2\times(4+1)^2}{4}\) \(\bigg(\because\,\displaystyle\sum_{k=1}^{n} k^3=\frac{n^2(n+1)^2}{4}\bigg)\) = \(\frac{81\times100}{4}\) = \(\frac{16\times16\times25}{4}\) = 2025 - 1600 = 425.

What is the sum of the series 13 – 23 + 33 – 43 + ........ + 93 equal to?

Answer»

Reqd. Sum = (1³ + 2³ + 3³ + 4³ + ..... + 9³) – 2(2³ + 4³ + 6³ + 8³)

= (1³+ 2³ + 3³ + ..... + 9³) – 2⁴ (1³ + 2³ + 3³ + 4³)

= \(\displaystyle\sum_{k=1}^{9} k^3\) - 2⁴ \(\displaystyle\sum_{k=1}^{4} k^3\) = \(\frac{9^2(9+1)^2}{4}\) - 2⁴ x \(\frac{4^2\times(4+1)^2}{4}\)

\(\bigg(\because\,\displaystyle\sum_{k=1}^{n} k^3=\frac{n^2(n+1)^2}{4}\bigg)\)

= \(\frac{81\times100}{4}\) = \(\frac{16\times16\times25}{4}\) = 2025 - 1600 = 425.