Find the sum of n terms of the series 1 + (1 + x) + (1 + x + x2) + ...

1.	Find the sum of n terms of the series 1 + (1 + x) + (1 + x + x2) + .....?
Answer» 1 + (1 + x) + (1 + x + x²) + ....... n terms ⇒ Required sum = \(\frac{1}{(1-x)}\) [(1 – x) + (1 – x) (1 + x) + (1 – x) (1 + x + x²) + ..... n terms] = \(\frac{1}{(1-x)}\) [(1 – x) + (1 – x²) + (1 – x³) + ..... n terms] = \(\frac{1}{(1-x)}\) [(1 + 1 + 1 + .... n terms) – (x + x² + x³ + ..... n terms] = \(\frac{1}{(1-x)}\)\(\bigg[n-\frac{x(1-x^n)}{(1-x)}\bigg]\) \(\bigg(\because{S_n}=\frac{a(1-r^n)}{(1-r)},\text{Here}\,a= x, r=x\bigg)\) \(\frac{n(1-x)-x(1-x^n)}{(1-x)^2}\).

Find the sum of n terms of the series 1 + (1 + x) + (1 + x + x2) + .....?

Answer»

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

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

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

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

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

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