# Find 10 partial sums of the series. (Round your answers to five decimal places.) \(\sum_{n=1}^{\infty }\frac{16}{(-3)^{n}}\)

**Solution:**

Given, \(\sum_{n=1}^{\infty }\frac{16}{(-3)^{n}}\)

We have to find the 10 partial sums of the series.

The sequence of terms is

When n= 1, 16/-3^{1}=16/-3=-5.33333

When n= 2, 16/-3^{2} = 16/9 =1.77778

When n= 3, 16/-3^{3 }= 16/-27=-0.59259

When n= 4, 16/-3^{4 }= 16/81 =0.19753

When n= 5, 16/-3^{5}= 16/-243 =-0.06584

When n= 6, 16/-3^{6}= 16/729 =0.02194

When n= 7,16/-3^{7}= 16/-2187=-0.00732

When n= 8, 16/-3^{8}= 16/6561=0.00244

When n= 9, 16/-3^{9}= 16/-19683 =-0.00081

When n= 10, 16/-3^{10}= 16/59049=0.00027

The partial sums are

S_{1}=16/-3^{ }=-5.33333

S_{2}=16/-3 +16/9=-5.33333+1.77778

=-3.55552

S_{3}=16/-3 +16/9+16/-27

=-3.55552-0.59259 =-4.14811

S_{4}=16/-3 +16/9+16/-27 + 16/81

=-4.14811+0.19753=-3.95058

S_{5}=16/-3 +16/9+16/-27 + 16/81 +16/-243

=-3.95058-0.06584=-4.01642

S_{6}=16/-3 +16/9+16/-27 + 16/81 +16/-243 +16/729

=-4.01642+0.02194=-3.99448

S_{7}=16/-3 +16/9+16/-27 + 16/81 +16/-243 +16/729 + 16/-2187

=-3.99448-0.00732=-4.00180

S_{8}=16/-3 +16/9+16/-27 + 16/81 +16/-243 +16/729 + 16/-2187 + 16/6561

=-4.00180+0.00244=-3.99936

S_{9}=16/-3 +16/9+16/-27 + 16/81 +16/-243 +16/729 + 16/-2187 + 16/6561 + 16/-19683

=-3.99936-0.00081=-4.00017

S_{10}=16/-3 +16/9+16/-27 + 16/81 +16/-243 +16/729 + 16/-2187 + 16/6561 + 16/-19683 + 16/59049

=-4.00017+0.00027=-3.99990

Therefore, the 10 partial sums are -5.33333, -3.55552, -4.14811, -3.95058, -4.01642, -3.99448, -4.00180, -3.99936, -4.00017 and -3.99990

## Find 10 partial sums of the series. (Round your answers to five decimal places.) \(\sum_{n=1}^{\infty }\frac{16}{(-3)^{n}}\)

**Summary:**

The 10 partial sums of the series. (Round your answers to five decimal places.) \(\sum_{n=1}^{\infty }\frac{16}{(-3)^{n}}\) are -5.33333, -3.55552, -4.14811, -3.95058, -4.01642, -3.99448, -4.00180, -3.99936, -4.00017 and -3.99990.

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