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प्रश्न
Find (502 – 492) + (482 –472) + (462 – 452) + .. + (22 –12).
उत्तर
(502 – 492) + (482 –472) + (462 – 452) + .. + (22 –12).
= (502 + 482 + 462 + ... + 22) – (492 + 472 + 452 + ... + 12)
= \[\displaystyle\sum_{r=1}^{25}(2r)^2 - \displaystyle\sum_{r=1}^{25}(2r - 1)^2\]
= \[\displaystyle\sum_{r=1}^{25} 4r^2 - \displaystyle\sum_{r=1}^{25} (4r^2 - 4r + 1)\]
= \[\displaystyle\sum_{r=1}^{25}[4r^2 - (4r^2 - 4r + 1)]\]
= \[\displaystyle\sum_{r=1}^{25}(4r - 1)\]
= 4\[\displaystyle\sum_{r=1}^{n}r - \displaystyle\sum_{r=1}^{n}1\]
= `4 xx (25(25 + 1))/2 - 25`
= `(4(25)(26))/2 - 25`
= 1300 – 25 = 1275.
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