Question Video: Simplifying Expansions Using the Binomial Theorem

Simplify 𝑛𝐢0 + 17 Γ— 𝑛𝐢1 + 17Β² Γ— 𝑛𝐢2+ ... + 17^(π‘Ÿ) Γ— π‘›πΆπ‘Ÿ + ... + 17 ^(𝑛) Γ— 𝑛𝐢𝑛.

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Video Transcript

Simplify 𝑛 𝐢 zero plus 17 multiplied by 𝑛 𝐢 one plus 17 squared multiplied by 𝑛 𝐢 two and so on plus 17 to the power of π‘Ÿ multiplied by π‘›πΆπ‘Ÿ and so on up to 17 to the power of 𝑛 multiplied by 𝑛𝐢𝑛.

In order to simplify this expression, we begin by recalling the binomial expansion of π‘Ž plus 𝑏 to the 𝑛th power. The first three terms of this expansion are 𝑛 𝐢 zero multiplied by π‘Ž to the power of 𝑛 plus 𝑛 𝐢 one multiplied by π‘Ž to the power of 𝑛 minus one multiplied by 𝑏 plus 𝑛 𝐢 two multiplied by π‘Ž to the power of 𝑛 minus two multiplied by 𝑏 squared. And the final term is 𝑛𝐢𝑛 multiplied by 𝑏 to the 𝑛th power. We notice that much of this is the same as the expression in this question. Instead of 𝑏 to the 𝑛th power, our last term contains 17 to the 𝑛th power. The second term contains 17 instead of 𝑏 and the third term, 17 squared instead of 𝑏 squared. This means that the value of 𝑏 is 17.

We notice that there is no π‘Ž part to any of our terms. Since one raised to any power is equal to one, we can assume that π‘Ž is equal to one as this is the only value for which this holds. The expression in the question is therefore equal to one plus 17 all raised to the 𝑛th power. And as one plus 17 equals 18, this is equal to 18 to the 𝑛th power.

The expression 𝑛 𝐢 zero plus 17 multiplied by 𝑛 𝐢 one plus 17 squared multiplied by 𝑛 𝐢 two and so on up to 17 to the 𝑛th power multiplied by 𝑛𝐢𝑛 is equal to 18 to the 𝑛th power.

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