#### To determine

**To find:** The value of *c*.

#### Answer

The value of c=3−12.

#### Explanation

**Given:**

The series is ∑n=2∞(1+c)−n=2.

**Result used:**

The sum of the geometric series (∑n=1∞arn−1 (or) a+ar+ar2+⋯) is a1−r if |r|<1, where *a* is the first term and *r* is the common ratio of the series.

Consider the series ∑n=2∞(1+c)−n.

Rewrite the above series and express it as follows.

∑n=2∞(1+c)−n=∑n=2∞1(1+c)n=1(1+c)2+1(1+c)3+1(1+c)4+⋯

Clearly, it is geometric series with the first term of the series is a=1(1+c)2 and the common ratio is r=11+c.

Use the Result stated above, the sum of the series ∑n=2∞(1+c)−n is computed as follows.

∑n=2∞(1+c)−n=1(1+c)21−11+c=1(1+c)21+c1+c−11+c=1(1+c)21+c−1(1+c)=1(1+c)2c(1+c)

Divide fraction and simplify the terms,

∑n=2∞(1+c)−n=1(1+c)2×(1+c)c=1(1+c)c=1c2+c

Since ∑n=2∞(1+c)−n=2, 1c2+c=2.

Cross multiply and simplify the terms,

2(c2+c)=12c2+2c=12c2+2c−1=0

Use quadratic formula and obtain the value of *c.*

c=−2±(2)2−4⋅2⋅(−1)2⋅2=−2±124=−2±234=−1±32

Note that, the geometric series is converges if |r|<1.

|11+c|<11|1+c|<1|1+c|>1

Here, |1+c|>1 if c=−1−32.

Therefore, the value of c=3−12.