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If a1,a2,a3,------an are in AP then prove that a12-a22+a32-a42+-------+a2n-1 2-a2n2=n/2n-1(a12-a2n 2)

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Given a1,a2,a3 are in AP Let d be the common difference and let d>1It means that a1-a2=a2-a3=a3-a4=.................a2n-1-a2n=-dNow let S a12-a22+a32-a42.............+a2n-12-a2n2S=(a1+a2)(a1-a2)+(a3+a4)(a3-a4)..........+(a2n-1+a2n)(a2n-1-a2n)S=(a1+a2)(-d)+(a3+a4)(-d)........+(a2n-1+a2n)(-d)S=(-d)a1+a2+a3+a4+............a2n-1+a2nS=(-d)2n2(a1+a2n)     sum of 2n terms of an A.P = 2n2×(1st term+last term)S=(-d)n(a1+a2n)Now since we can write a1-a2n=(a1-a2)+(a2-a3)+.........(a2n-1-a2n)=(2n-1)(-d)       Hence -d =a1-a2n(2n-1)Now replacing -d by a1-a2n(2n-1)in S=(-d)n(a1+a2n),  we getS=a1-a2n(2n-1)n(a1+a2n)=n2n-1(a12-a2n2)   ∴ a12-a22+a32-a42.............+a2n-12-a2n2=n2n-1(a12-a2n2)  proved

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