# What do I do to implement the #x^2# into this series? #x^2sum_(n=0)^oo(na_nx^(n-1))#

# sum_(n=0)^oo (na_nx^(n+1))#

If unclear as to the effect then the best option to expand a few terms of the summation:

Then we can put it the series back into "sigma" notation:

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To implement ( x^2 ) into the given series ( \sum_{n=0}^{\infty} (na_nx^{n-1}) ), you need to differentiate the series with respect to ( x ) twice. This will shift the index of summation, ( n ), by ( -1 ), and multiply each term by ( n ), effectively introducing ( x^2 ) into the series.

The process involves differentiating term by term and then summing the resulting series. After the first differentiation, you will have ( \sum_{n=0}^{\infty} (na_nx^{n-2}) ), and after the second differentiation, you will have ( \sum_{n=0}^{\infty} (n(n-1)a_nx^{n-3}) ).

So, to implement ( x^2 ) into the series, differentiate the series twice with respect to ( x ), and then sum the resulting series.

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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