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प्रश्न
Find (702 – 692) + (682 – 672) + ... + (22 – 12)
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उत्तर
Let S = (702 – 692) + (682 – 672) + ... + (22 – 12)
∴ S = (22 – 12) + (42 – 32) + ... + (702 – 692)
Here, 2, 4, 6, …, 70 is an A.P. with rth term = 2r
and 1, 3, 5, …, 69 in A.P. with rth term = 2r – 1
∴ S = \[\displaystyle\sum_{r=1}^{35} [(2r)^2 - (2r - 1)^2]\]
= \[\displaystyle\sum_{r=1}^{35}[4r^2 - (4r^2 - 4r + 1)]\]
= \[\displaystyle\sum_{r=1}^{35}(4r - 1)\]
= 4 \[\displaystyle\sum_{r=1}^{35} r - \displaystyle\sum_{r=1}^{35} 1\]
= `4.(35 xx 36)/2 - 35`
= (72 – 1) (35)
= 71 x 35 = 2485.
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