Question

12) Find the sum, S7, for the geometric series (1 + 5 + 25+ --- + 15625).

Answer 1

Answer:

Find the (r) between adjacent members

a2/a1=+25/5=5

a3/a2=+125/+25=5

a4/a3=+625/+125=5

a5/a4=+3125/+625=5

a6/a5=+15625/+3125=5

The ration (r) between every two adjacent members of the series is constant and equal to 5

General Form: a

n

=a

1

×r

n-1

This geometric series: a

n

=5×5

n-1

The nature of this series

Sum of finite geometric series members

sum of a geometric series

The sum of our particular series

a 1-rn = 5 1-56 = 5 1-15625 = 5 -15624 =5×3906=19530

1-r 1-5 -4 -4

Finding the n

th

element

a2 =a1×r2-1  =a1×r1 =5×51 =25

a3 =a1×r3-1  =a1×r2 =5×52 =125

a4 =a1×r4-1  =a1×r3 =5×53 =625

a5 =a1×r5-1  =a1×r4 =5×54 =3125

a6 =a1×r6-1  =a1×r5 =5×55 =15625

a7 =a1×r7-1  =a1×r6 =5×56 =78125

a8 =a1×r8-1  =a1×r7 =5×57 =390625

a9 =a1×r9-1  =a1×r8 =5×58 =1953125

a10 =a1×r10-1  =a1×r9 =5×59 =9765625

a11 =a1×r11-1  =a1×r10 =5×510 =48828125

a12 =a1×r12-1  =a1×r11 =5×511 =244140625

a13 =a1×r13-1  =a1×r12 =5×512 =1220703125

a14 =a1×r14-1  =a1×r13 =5×513 =6103515625

a15 =a1×r15-1  =a1×r14 =5×514 =30517578125

a16 =a1×r16-1  =a1×r15 =5×515 =152587890625