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Statement-1: The sum of n terms of the s...

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.

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a+(a+d)+(a+2d)+....+[a+(n-1)d]=(n)/(2)(2a+(n-1)d)

Sum of n terms of the series (2n-1)+2(2n-3)+3(2n-5)+... is

Statement 1 The sum of first n terms of the series 1^(2)-2^(2)+3^(2)-4^(2)_5^(2)-"……" can be =+-(n(n+1))/(2) . Statement 2 Sum of first n narural numbers is (n(n+1))/(2)

Theorem: The sum of nth terms of an AP with first term a and common difference d is S_(n)=(n)/(2)(2a+(n-1)d)

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]

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

Show that sum S_(n) of n terms of an AP with first term a and common difference d is S_(n)=(n)/(2)(2a+(n-1)d)

Knowledge Check

  • The sum of (2n+1) terms of the series a-(a+d)+a(a+ad)-(a+3d)+… is :

    A
    a+3nd
    B
    3a+nd
    C
    2a+3nd
    D
    a+nd

  • The sum upto (2n+1) terms of the series a^(2)-(a+d)^(2)+(a+2d)^(2)-(a+3d)^(2)+… is

    A
    `a^(2)+3nd^(2)`
    B
    `a^(2)+2nad+n(n-1)d^(2)`
    C
    `a^(2)+3nad+n(n-1)d^(2)`
    D
    `(a+nd)^(2)+n(n+1)d^(2)`

  • Statement 1 The sum of first n terms of the series 1^(2)-2^(2)+3^(2)-4^(2)_5^(2)-"……" can be =+-(n(n+1))/(2) . Statement 2 Sum of first n narural numbers is (n(n+1))/(2)

    A
    Statement 1 is true, Statement 2 is true, Statement 2 is a corrct explanation for Statement 1.
    B
    Statement 1 is true, Statement 2 is true, Statement 2 is not a correct explanation for Statement 1.
    C
    Statement 1 is true, Statement 2 is false.
    D
    Statement 1 is false, Statement 2 is true.