Maharashtra State BoardHSC Commerce 11th
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Find (502 – 492) + (482 –472) + (462 – 452) + .. + (22 –12). - Mathematics and Statistics

Sum

Find (502 – 492) + (482 –472) + (462 – 452) + .. + (22 –12).

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Solution

(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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APPEARS IN

Balbharati Mathematics and Statistics 1 (Commerce) 11th Standard Maharashtra State Board
Chapter 4 Sequences and Series
Miscellaneous Exercise 4 | Q 16 | Page 64
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