Consider the sum S = a + a𝑤 + a𝑤² +···+ a𝑤^n−1 = ∑i=1^n a𝑤^i−1 of the first n terms in a geometric series. Show that the sum S in the above formula equals ∑i=1^n a𝑤^i−1 = a(1 − 𝑤^n)/1 − 𝑤, provided that 𝑤 ≠ 1. (For 𝑤 = 1, we have trivially S = na.)

Question

Consider the sum

[latex]S=a+a w+a w^{2}+\cdot\cdot\cdot\cdot+a w^{n-1}=\sum_{i=1}^{n}a w^{i-1}[/latex]

of the first n terms in a geometric series. Show that the sum S in the above formula equal

[latex]\sum_{i=1}^{n}a w^{i-1}={\frac{a(1-w^{n})}{1-w}},[/latex]

provided that 𝑤 ≠ 1. (For 𝑤 = 1, we have trivially S = na.)

Accepted Answer

Multiplying each term of S by 𝑤 we obtain that

[latex]w S=a w+a w^{2}+a w^{3}+\cdot\cdot\cdot+a w^{n}.[/latex]

Therefore,

[latex]w S-S=(a w+a w^{2}+a w^{3}+\cdot\cdot\cdot+a w^{n})-(a+a w+a w^{2}+\cdot\cdot\cdot+a w^{n-1})= a 𝑤^n − a[/latex]

and for 𝑤 ≠ 1 this gives immediately

[latex]S={\frac{a(w^{n}-1)}{w-1}}={\frac{a(1-w^{n})}{1-w}}.[/latex]

The following result is particularly useful.

Question A.2: Consider the sum S = a + a𝑤 + a𝑤² +···+ a𝑤^n−1 = ∑i=1^n a𝑤^i......