To find:

The sum of the infinite series \(\sum_{i=1}^\infty12(-0.7)^{i-1}\)

Concept used:

The sum of infinite geometric series can be obtained by the formula,

\(S_\infty=\frac{a}{1-r}\quad(r<1)\)</span>

Here, a is first term, r is common ratio and \(S_\infty\) is sum of infinite geometric series.

Calculation:

Expand the summation as follows:

\(\sum_{i=1}^\infty12(-0.7)^{i-1}=12(-0.7)^{1-1}+12(-0.7)^{2-1}+12(-0.7)^{3-1}+...\)

\(=12+12(-0.7)+12(-0.7)^2+...\)

It is observed from the series that the first term is 12 and common ratio is (−0.7) and common ratio is less than 1.

Substitute 12 for a and (−0.7) for r in equation (1).

\(S_{\infty}=\frac{12}{10(-0.7)}\)

\(=\frac{12}{1+0.7}\)

\(=\frac{12}{1.7}\)

\(=7.06\)

Thus, the sum of the infinite series is 7.06.

The sum of the infinite series \(\sum_{i=1}^\infty12(-0.7)^{i-1}\)

Concept used:

The sum of infinite geometric series can be obtained by the formula,

\(S_\infty=\frac{a}{1-r}\quad(r<1)\)</span>

Here, a is first term, r is common ratio and \(S_\infty\) is sum of infinite geometric series.

Calculation:

Expand the summation as follows:

\(\sum_{i=1}^\infty12(-0.7)^{i-1}=12(-0.7)^{1-1}+12(-0.7)^{2-1}+12(-0.7)^{3-1}+...\)

\(=12+12(-0.7)+12(-0.7)^2+...\)

It is observed from the series that the first term is 12 and common ratio is (−0.7) and common ratio is less than 1.

Substitute 12 for a and (−0.7) for r in equation (1).

\(S_{\infty}=\frac{12}{10(-0.7)}\)

\(=\frac{12}{1+0.7}\)

\(=\frac{12}{1.7}\)

\(=7.06\)

Thus, the sum of the infinite series is 7.06.