### Video Transcript

Find the sum of terms of the infinite geometric sequence starting at π sub one with πth term π sub π equals three multiplied by 14 to the power of one minus π.

Letβs begin by recalling the formula for the sum of terms of an infinite geometric sequence. Itβs π sub β equals π sub one over one minus π, where π sub one is the first term in the sequence and π is the common ratio. In order for this formula to be valid, the absolute value of the common ratio π must be strictly less than one, as this is necessary for the sequence to be convergent. In order to answer this question then, we need to determine the values of both the first term and the common ratio in this sequence.

Now, in general, the πth term or general term of a geometric sequence is given by π sub π is equal to π sub one multiplied by π to the power of π minus one. This looks similar to the structure of the πth term rule weβve been given. And so we may expect that the first term in this sequence is equal to three. But if we look carefully at the exponent of π, we see that in the general term the exponent is π minus one, whereas in the formula weβve been given, the exponent is one minus π, which is the negative of this.

Letβs see if we can manipulate the base of this exponential expression so that the exponent is π minus one. Essentially, we want to change the sign of the exponent, so we want to multiply it by negative one. We know that, in general, negative exponents define reciprocals. 14 is equal to one over 14, thatβs the reciprocal of 14, to the power of negative one.

So we could rewrite this exponential term as one over 14 to the power of negative one all to the power of one minus π. This is helpful because when we raise a value to an exponent and then to another exponent, we multiply the exponents together. π₯ to the power of π¦ to the power of π§ is π₯ to the power of π¦π§. So one over 14 to the power of negative one to the power of one minus π is one over 14 to the power of negative one times one minus π, which is one over 14 to the power of π minus one.

The general term of this sequence then can be written as three multiplied by one over 14 to the power of π minus one. And itβs now in exactly the same form as the general term. We see that the first term is equal to three and the common ratio is one over 14. We can now substitute these values of π sub one and π into the infinite sum formula. As the value of π is one over 14, its absolute value is less than one. And so this sum does exist. We have that π sub β is equal to three over one minus one over 14. Thatβs three divided by 13 over 14. And then we can flip this fraction and multiply. So we have three multiplied by 14 over 13, which is 42 over 13.

We could write this as a mixed number, or we can leave it as an improper fraction. And it canβt be simplified as there are no common factors in the numerator and denominator other than one. We found then that the sum of the terms of the infinite geometric sequence with the given πth term rule is 42 over 13.