Math, asked by Anonymous, 19 days ago

Please answer my question asap​

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Answered by Anonymous
4

Answer:

In order to solve the given problem, we must know the formula for finding the sum of squares of first n natural numbers.

We can see that in the numerator of the given problem, it's a series of sum of squares of n natural number.

We have a formula:

  •  \boxed{ \sum \limits_{r = 1}^n {r}^{2}  =  \dfrac{n(n + 1)(2n + 1)}{6} }

By Using this formula in the given limit, we get:

 \implies \lim \limits_{x \to  \infty} \dfrac{1 + 4 + 9 + ... +  {x}^{2} }{ {x}^{3} }

 \implies \lim \limits_{x \to  \infty} \dfrac{x(x+ 1)(2x + 1)}{6 {x}^{3}}

 \implies \lim \limits_{x \to  \infty} \dfrac{(x+ 1)(2x + 1)}{6 {x}^{2}}

{ \implies \lim \limits_{x \to  \infty} \dfrac{x(2x + 1) + 1(2x + 1)}{6 {x}^{2}}}

{ \implies \lim \limits_{x \to  \infty} \dfrac{2{x}^{2} + x  +2x + 1}{6 {x}^{2}}}

{ \implies \lim \limits_{x \to  \infty} \dfrac{2{x}^{2} + 3x + 1}{6 {x}^{2}}}

{ \implies \lim \limits_{x \to  \infty} \dfrac{ {x}^{2} \left[ 2 +  \dfrac{3}{x}  +  \dfrac{1}{ {x}^{2} } \right]}{6 {x}^{2}}}

{ \implies \lim \limits_{x \to  \infty} \dfrac{ 2 +  \dfrac{3}{x}  +  \dfrac{1}{ {x}^{2} }}{6}}

Now Substitute the limits:

{ \implies \dfrac{ 2 +  \dfrac{3}{ \infty}  +  \dfrac{1}{ { \infty}^{2} }}{6}}

{ \implies \dfrac{ 2 +  0 + 0}{6}}

{ \implies \dfrac{1}{3}}

So our required answer is:

 \underline{ \boxed{\lim \limits_{x \to  \infty} \dfrac{1 + 4 + 9 + ... +  {x}^{2} }{ {x}^{3} }  =  \frac{1}{3}} }

Answered by honey8315
0

Answer:

1 =1³

2=2³

3=3³

and so on

answer is 1/4

Step-by-step explanation:

hope it helps mark me brainliest

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