The sum of n terms of the following series 1 + (1 + x) + (1 + x + x2)

1.	The sum of n terms of the following series 1 + (1 + x) + (1 + x + x2) + .... will be(a) \(\frac{1-x^n}{1-x}\)(b) \(\frac{x(1-x^n)}{1-x}\)(c) \(\frac{n(1-x)-x(1-x^n)}{(1-x)^2}\)(d) None of the above
Answer» (c) \(\frac{n(1-x)-x(1-x^n)}{(1-x)^2}\) S_n = 1 + (1 + x) + (1 + x + x²) + .... n terms ⇒ Sn = \(\frac{1}{1-x}\) [(1 – x) + (1 – x) (1 + x) + (1 – x) (1 + x + x²) + .... to n terms] = \(\frac{1}{1-x}\) [(1 – x) + (1 – x²) + (1 – x³) + .... to n terms] = \(\frac{1}{1-x}\) [n – (x + x² + x⁴ + .... to n terms] = \(\frac{1}{1-x}\) \(\bigg[n-\frac{x(1-x^n)}{1-x}\bigg]\) \(\bigg(\because\,S_n = \frac{a(1-r^n)}{1-r}\bigg)\) = \(\frac{n(1-x)-x(1-x^n)}{(1-x)^2}\)

The sum of n terms of the following series 1 + (1 + x) + (1 + x + x2) + .... will be(a) \(\frac{1-x^n}{1-x}\)(b) \(\frac{x(1-x^n)}{1-x}\)(c) \(\frac{n(1-x)-x(1-x^n)}{(1-x)^2}\)(d) None of the above

Answer»

(c) \(\frac{n(1-x)-x(1-x^n)}{(1-x)^2}\)

S_n = 1 + (1 + x) + (1 + x + x²) + .... n terms

⇒ Sn = \(\frac{1}{1-x}\) [(1 – x) + (1 – x) (1 + x) + (1 – x) (1 + x + x²) + .... to n terms]

= \(\frac{1}{1-x}\) [(1 – x) + (1 – x²) + (1 – x³) + .... to n terms]

= \(\frac{1}{1-x}\) [n – (x + x² + x⁴ + .... to n terms]

= \(\frac{1}{1-x}\) \(\bigg[n-\frac{x(1-x^n)}{1-x}\bigg]\) \(\bigg(\because\,S_n = \frac{a(1-r^n)}{1-r}\bigg)\)

= \(\frac{n(1-x)-x(1-x^n)}{(1-x)^2}\)