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Can anyone tell me how to calculate general sum of sequence till \(n\) . \[1^{4}+2^{4}+3^{4}+ \ldots +n^{4}\]

Note by Akshat Sharda 2 years, 2 months ago

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Hint:

\[ \begin{eqnarray} \sum_{k=1}^n \left( (k+1)^2 - k^2 \right) &= &2 \sum_{k=1}^n k + \sum_{k=1}^n 1 \\ \displaystyle (n+1)^2 - 1 &=& 2 \color{blue}{\sum_{k=1}^n k} + n \\ \displaystyle \sum_{k=1}^n k &=& \frac{n(n+1)}2 \end{eqnarray} \]

And

\[ \begin{eqnarray} \sum_{k=1}^n \left( (k+1)^3 - k^3 \right) &= & \sum_{k=1}^n (3k^2 + 3k + 1) \\ \displaystyle (n+1)^3 - 1 &=& 3\color{blue}{ \sum_{k=1}^n k^2} +3 \sum_{k=1}^n k + \sum_{k=1}^n 1 \\ \displaystyle (n+1)^3 - 1 &=& 3\color{blue}{ \sum_{k=1}^n k^2} +3\frac{n(n+1)}2 + n \\ \sum_{k=1}^n k^2 &=& \frac16 n(n+1)(2n+1) \\ \end{eqnarray} \]

Can you find a general pattern to it?

Answer is \( \frac1{30} n(n+1)(2n+1)(3n^2 + 3n-1) \). You can double check your work by induction. – Pi Han Goh · 2 years, 2 months ago

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@Pi Han Goh – I appriciate what you did and Thank you for the same. – Akshat Sharda · 2 years, 2 months ago

@Pi Han Goh – I have a question- Will this pattern work to find general sum of any power of \(n\)? – Akshat Sharda · 2 years, 2 months ago

@Akshat Sharda – Study Bernoulli's Numbers. You'll get the answer for nth power sums. :) – Satyajit Mohanty · 2 years, 2 months ago

@Satyajit Mohanty – I find that more troublesome, unless you got a table for the number of Bernoulli numbers, then yes, it will be easier, otherwise it takes up more time. – Pi Han Goh · 2 years, 2 months ago

@Pi Han Goh – Well, it's true. I was just giving a general idea! :) – Satyajit Mohanty · 2 years, 2 months ago

@Akshat Sharda – For positive integer powers, yes. – Pi Han Goh · 2 years, 2 months ago

Study Faulhaber's Formula for a General Idea. Also visit Bernoulli's numbers if it interests you. :) – Satyajit Mohanty · 2 years, 2 months ago

1/30(n+1)(2n+1)(3n^2+3n-1) – Aakash Khandelwal · 2 years, 2 months ago

(6x + 15x + 10x - x) / 30 – Aashirvad Raj · 2 years, 2 months ago

@Aashirvad Raj – Hm, that simplifies to just \(x\). Are you missing some exponents? – Calvin Lin Staff · 2 years, 2 months ago

@Aashirvad Raj – Even I know the formula but how to derive it is much more important. – Akshat Sharda · 2 years, 2 months ago

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## Comments

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TopNewestHint:\[ \begin{eqnarray} \sum_{k=1}^n \left( (k+1)^2 - k^2 \right) &= &2 \sum_{k=1}^n k + \sum_{k=1}^n 1 \\ \displaystyle (n+1)^2 - 1 &=& 2 \color{blue}{\sum_{k=1}^n k} + n \\ \displaystyle \sum_{k=1}^n k &=& \frac{n(n+1)}2 \end{eqnarray} \]

And

\[ \begin{eqnarray} \sum_{k=1}^n \left( (k+1)^3 - k^3 \right) &= & \sum_{k=1}^n (3k^2 + 3k + 1) \\ \displaystyle (n+1)^3 - 1 &=& 3\color{blue}{ \sum_{k=1}^n k^2} +3 \sum_{k=1}^n k + \sum_{k=1}^n 1 \\ \displaystyle (n+1)^3 - 1 &=& 3\color{blue}{ \sum_{k=1}^n k^2} +3\frac{n(n+1)}2 + n \\ \sum_{k=1}^n k^2 &=& \frac16 n(n+1)(2n+1) \\ \end{eqnarray} \]

Can you find a general pattern to it?

Answer is \( \frac1{30} n(n+1)(2n+1)(3n^2 + 3n-1) \). You can double check your work by induction. – Pi Han Goh · 2 years, 2 months ago

Log in to reply

– Akshat Sharda · 2 years, 2 months ago

I appriciate what you did and Thank you for the same.Log in to reply

– Akshat Sharda · 2 years, 2 months ago

I have a question- Will this pattern work to find general sum of any power of \(n\)?Log in to reply

– Satyajit Mohanty · 2 years, 2 months ago

Study Bernoulli's Numbers. You'll get the answer for nth power sums. :)Log in to reply

– Pi Han Goh · 2 years, 2 months ago

I find that more troublesome, unless you got a table for the number of Bernoulli numbers, then yes, it will be easier, otherwise it takes up more time.Log in to reply

– Satyajit Mohanty · 2 years, 2 months ago

Well, it's true. I was just giving a general idea! :)Log in to reply

– Pi Han Goh · 2 years, 2 months ago

For positive integer powers, yes.Log in to reply

Study Faulhaber's Formula for a General Idea. Also visit Bernoulli's numbers if it interests you. :) – Satyajit Mohanty · 2 years, 2 months ago

Log in to reply

1/30(n+1)(2n+1)(3n^2+3n-1) – Aakash Khandelwal · 2 years, 2 months ago

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(6x + 15x + 10x - x) / 30 – Aashirvad Raj · 2 years, 2 months ago

Log in to reply

– Calvin Lin Staff · 2 years, 2 months ago

Hm, that simplifies to just \(x\). Are you missing some exponents?Log in to reply

– Akshat Sharda · 2 years, 2 months ago

Even I know the formula but how to derive it is much more important.Log in to reply