Statement-1: `1+2+3....+n=(n(n+1))/(2),"for all "n in N`
Statement-2: `a+(a+d)+(a+2d)+....+(a+(n-1)d)=(n)/(2)[2a+(n-1)d]`

Statement-1: 1^(2)+2^(2)+....+n^(2)=(n(n+1)(2n+1))/(6)"for all "n in N Statement-2: 1+2+3....+n=(n(n+1))/(2),"for all"n in N

Statement-1: 1^(2)+2^(2)+....+n^(2)=(n(n+1)(2n+1))/(6)"for all "n in N Statement-2: 1+2+3....+n=(n(n+1))/(2),"for all"n in N

By the Principle of Mathematical Induction, prove the following for all n in N : a+ (a+d)+(a+2d)+....+ [a+(n-1)d] =n/2 [2a + (n-1)d] .

Statement-1: The sum of n terms of the series a+(a+d)+(a+2d)+...(a+(n-1)d)=(n)/(2)[2a+(n-1)d] Statement-2:- Mathematical induction is valid only for natural numbers.

Statement-1: The sum of n terms of the series a+(a+d)+(a+2d)+...(a+(n-1)d)=(n)/(2)[2n+(n-1)d] Statement-2:- Mathematical induction is valid only for natural numbers.

Prove the following by using the principle of mathematical induction for all n in N a + (a+d)+(a+2d) +……+ (a+(n-1)d) = n/2 [2a + (n-1)d]

Prove that a+(a+d)+(a+2d)+........n "terms"=n/2(2a+(n-1)d)

Prove the following by the principle of mathematical induction: a+(a+d)+(a+2d)++(a+(n-1)d)=(n)/(2)[2a+(n-1)d]

Prove the following by the principle of mathematical induction: a+(a+d)+(a+2d)++(a+(n-1)d)=n/2[2a+(n-1)d]