Question

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

Answer 1

(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.