Evaluate:\(\displaystyle\sum_{r=1}^{10} (2r-1)^2\)

1.	Evaluate:\(\displaystyle\sum_{r=1}^{10} (2r-1)^2\)
Answer» \(\displaystyle\sum_{i=1}^{10} (2r-1)^2\) = 1² + 3² + ⋯ + 17² + 19² To find the sum of above 10 terms, we find the sum of n terms in general. ∴ Let T_n be n^th term of the series. 1² + 3² + 5²+ 7² + ..... to n terms Now, T_n = [1 + (n − 1) × 2]² = 4n² − 4n + 1 ∴ 1² + 3² + 5² + ⋯ to n terms = \(\displaystyle\sum_{k=1}^{n} T_k\) = \(\displaystyle\sum_{k=1}^{n} (4k^2-4k'1)\) = \(4\bigg(\displaystyle\sum_{k=1}^{n} k^2\bigg)-4\bigg(\displaystyle\sum_{k=1}^{n} k\bigg)+\) \(\bigg(\displaystyle\sum_{k=1}^{n} 1\bigg)\) = 4 \(\bigg[\frac{n(n + 1)(2n + 1)}{6}\bigg]\) − 4 \(\bigg[\frac{n(n + 1)}{6}\bigg]+n\) = \(\frac{n}{3}\)[2(n + 1)(2n + 1) − 6(n + 1) + 3] = \(\frac{n}{3}\)[4n² + 6n + 2 − 6n − 6 + 3] = \(\frac{n}{3}\)(4n² − 1) Now put n = 10 in the above result, we get 1² + 3² + 5² + ⋯ + 17² + 19² = \(\frac{10}{3}\) (4.10² − 1) = \(\frac{10}{3}\)(399) = 1330

Answer»

\(\displaystyle\sum_{i=1}^{10} (2r-1)^2\) = 1² + 3² + ⋯ + 17² + 19²

To find the sum of above 10 terms, we find the sum of n terms in general.

∴ Let T_n be n^th term of the series.

1² + 3² + 5²+ 7² + ..... to n terms

Now, T_n = [1 + (n − 1) × 2]² = 4n² − 4n + 1

∴ 1² + 3² + 5² + ⋯ to n terms

= \(\displaystyle\sum_{k=1}^{n} T_k\)

= \(\displaystyle\sum_{k=1}^{n} (4k^2-4k'1)\)

= \(4\bigg(\displaystyle\sum_{k=1}^{n} k^2\bigg)-4\bigg(\displaystyle\sum_{k=1}^{n} k\bigg)+\) \(\bigg(\displaystyle\sum_{k=1}^{n} 1\bigg)\)

= 4 \(\bigg[\frac{n(n + 1)(2n + 1)}{6}\bigg]\) − 4 \(\bigg[\frac{n(n + 1)}{6}\bigg]+n\)

= \(\frac{n}{3}\)[2(n + 1)(2n + 1) − 6(n + 1) + 3]

= \(\frac{n}{3}\)[4n² + 6n + 2 − 6n − 6 + 3]

= \(\frac{n}{3}\)(4n² − 1)

Now put n = 10 in the above result, we get

1² + 3² + 5² + ⋯ + 17² + 19² = \(\frac{10}{3}\) (4.10² − 1)

= \(\frac{10}{3}\)(399)

= 1330

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