If the sequence `a_(1),a_(2),a_(3),…,a_(n)` is an A.P., then prove that `a_(1)^(2)-a_(2)^(2)+a_(3)^(2)-a_(4)^(2)+…+a_(2n-1)^(2)-a_(2n)^(2)=n/(2n-1)(a_(1)^(2)-a_(2n)^(2))`

Explanation

Let d be the common difference of the A.P. Then,
`d=a_(2)-a_(1)=a_(3)-a_(2)=a_(4)-a_(3)=cdotcdotcdot=a_(2n)-a_(2n-1)`
Now, `a_(1)^(2)=a_(2)^(2)+a_(3)^(2)-a_(4)^(2)+…+a_(2n-1)^(2)-a_(2n)^(2)`
`=(a_(1)+a_(2))(a_(1)-a_(2))+(a_(3)+a_(4))(a_(3)-a_(4))+..+(a_(2n-1)+a_(2n))xx(a_(2n-1)-a_(2n))`
`=-d(a_(1)+a_(2)+a_(3)+...+a_(2n))`
`=-d(2n)/2(a_(1)+a_(2n))`
`-dn((a_(1)^(2)-a_(2n)^(2)))/(a_(1)-a_(2n))`
`=(dn(a_(1)^(2)-a_(2n)^(2)))/(a_(2n)-a_(1))`
`=n/(2n-1)(a_(1)^(2)-a_(2n)^(2))` [Using `a_(2n)=a_(1)+(2n-1)d`]