Question

Find (70² – 69²) + (68² – 67²) + ... + (2² – 1²)

Answer 1

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.