Question

the sum of n terms of an AP is an (n-1) . The sum of the squares of these terms is equal to

Answer 1

Sum to n terms: S₁ = a*n(n - 1) ------ (1) [Given]
So sum to (n - 1) terms of AP is: S₂ = a*(n - 1){(n - 1) - 1)} = a(n - 1)(n - 2) ----- (2)

Hence nth term of AP = S₁ - S₂ [Eq (1) - Eq(2)]:

= a*n(n - 1) - a(n - 1)(n - 2) = a(n - 1)(n - n + 2) = 2a(n - 1)

Substituting n = 1, 2, 3, 4, -------
the AP is: 0, 2a, 4a, 6a, 8a, 10a, -----------

ii) Hence of the above, the sequence whose terms are square of each term:

0, 4a², 16a², 36a², 64a², ----

Sum of these squares is: 0 + 4a² + 16a² + 36a² + 64a² + ---

S = 4a²[0 + 1 + 4 + 9 + 16 + ---]

= 4a²[0² + 1² + 2² + 3² + 4² + 5² + ---]

= 4a²[1² + 2² + 3² + 4² + ------ + (n - 1)²]
{Number of terms are only (n - 1) terms}

Sum of squares of first n natural numbers = n(n + 1)(2n + 1)/6
Hence sum to (n - 1) terms = (n - 1)(n)(2n - 1)/6

Thus sum of the squares S = 4a²(n)(n - 1)(2n - 1)/6

Answer 2

Hey !!

Let Sₙ = an(n-1), then

     Sₙ₋₁ = a(n-1)(n-2)

    ∴                   Tₙ = Sₙ - Sₙ₋₁ = 2a(n-1)

Tₙ² = 4a² (n-1)²

∴ Sum = ΣTₙ² = 4a² (n-1)(n)(2n-1) / 6

   = 2a² n(n-1)(2n-1) / 3

Good luck !!