Question

solve this sum
I am unable to do it
​

Answer 1

EXPLANATION.

Sum of the series.

⇒ √3 + 3√2 + 6√3 + . . . . . . to 16 terms.

As we know that,

It is a series of geometric progression.

First term = a = √3.

Common ratios = r = b/a.

⇒ 3√2/√3.

Rationalize the equation, we get.

⇒ 3√2/√3 x √3/√3. = 3√6/3 = √6.

Common ratios = r = b/a.

⇒ 6√3/3√2 = 2√3/√2.

Rationalize the equation, we get.

⇒ 2√3/√2 x √2/√2 = 2√6/2 = √6.

As we can see that,

⇒ Common ratios = r = b/a = √6.

Formula of Sₙ terms of G.P.

⇒ Sₙ = a(rⁿ - 1)/(r - 1).

⇒ S₁₆ = √3[(√6)¹⁶ - 1)]/(√6 - 1).

⇒ S₁₆ = √3[(6)^(16/2) - 1)]/(√6 - 1).

⇒ S₁₆ = √3[(6⁸ - 1)]/(√6 - 1).

                                                                                                                         

MORE INFORMATION.

General terms of a G.P.

General term (nth term) of a G.P. : a + ar + ar² + . . . . . . is given by,

Tₙ = arⁿ⁻¹.

Sum of n terms of a G.P.

The sum of first n terms of an G.P is given by,

Sₙ = a(1 - rⁿ)/(1 - r) = a - rTₙ/1 - r when r < 1.

Sₙ = a(rⁿ - 1)/(r - 1) = rTₙ - a/r - 1 when r > 1.

Sₙ = nr when r = 1.

Sum of an infinite G.P.

The sum of an infinite G.P. with first term a and common ratio r,

(- 1 < r < 1 that is |r| < 1) is S∞ = a/1 - r.