Maharashtra State BoardHSC Science (Electronics) 11th
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Answer the following: Find (502 – 492) + (482 – 472) + (462 – 452) + ... + (22 – 12) - Mathematics and Statistics

Sum

Answer the following:

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)

= `sum_("r" = 1)^25 (2"r")^2 - sum_("r" = 1)^25(2"r" - 1)^2`

= `sum_("r" = 1)^25 4"r"^2 - sum_("r" = 1)^25(4"r"^2 - 4"r" + 1)`

= `sum_("r" = 1)^25[4"r"^2 - (4"r"^2 - 4"r" + 1)]`

= `sum_("r" = 1)^25(4"r" - 1)`

= `4sum_("r" = 1)^25 "r" - sum_("r" = 1)^25 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 2 (Arts and Science) 11th Standard Maharashtra State Board
Chapter 2 Sequences and Series
Miscellaneous Exercise 2 | Q II. (19) | Page 42
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