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Sum of 9th Powers of First N Natural Numbers Formula

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The Sum of 9th Powers of First N Natural Numbers is the summation of the 9th powers of the natural numbers starting from 1 to the nth natural number. Check FAQs

S n9 = \frac{n^{2} \cdot (n^{2} + n - 1) \cdot (2 \cdot n^{4} + 4 \cdot n^{3} - n^{2} - 3 \cdot n + 3) \cdot {(n + 1)}^{2}}{20}

S_n9 - Sum of 9th Powers of First N Natural Numbers?n - Value of N?

Sum of 9th Powers of First N Natural Numbers Example

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Here is how the Sum of 9th Powers of First N Natural Numbers equation looks like with Values.

Here is how the Sum of 9th Powers of First N Natural Numbers equation looks like with Units.

Here is how the Sum of 9th Powers of First N Natural Numbers equation looks like.

20196 = \frac{3^{2} \cdot (3^{2} + 3 - 1) \cdot (2 \cdot 3^{4} + 4 \cdot 3^{3} - 3^{2} - 3 \cdot 3 + 3) \cdot {(3 + 1)}^{2}}{20}

20196 = \frac{3^{2} \cdot (3^{2} + 3 - 1) \cdot (2 \cdot 3^{4} + 4 \cdot 3^{3} - 3^{2} - 3 \cdot 3 + 3) \cdot {(3 + 1)}^{2}}{20}

20196 = \frac{3^{2} \cdot (3^{2} + 3 - 1) \cdot (2 \cdot 3^{4} + 4 \cdot 3^{3} - 3^{2} - 3 \cdot 3 + 3) \cdot {(3 + 1)}^{2}}{20}

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Sum of 9th Powers of First N Natural Numbers Solution

Follow our step by step solution on how to calculate Sum of 9th Powers of First N Natural Numbers?

FIRST Step Consider the formula

S n9 = \frac{n^{2} \cdot (n^{2} + n - 1) \cdot (2 \cdot n^{4} + 4 \cdot n^{3} - n^{2} - 3 \cdot n + 3) \cdot {(n + 1)}^{2}}{20}

Next Step Substitute values of Variables

∴

S n9 = \frac{3^{2} \cdot (3^{2} + 3 - 1) \cdot (2 \cdot 3^{4} + 4 \cdot 3^{3} - 3^{2} - 3 \cdot 3 + 3) \cdot {(3 + 1)}^{2}}{20}

Next Step Prepare to Evaluate

∴

S n9 = \frac{3^{2} \cdot (3^{2} + 3 - 1) \cdot (2 \cdot 3^{4} + 4 \cdot 3^{3} - 3^{2} - 3 \cdot 3 + 3) \cdot {(3 + 1)}^{2}}{20}

LAST Step Evaluate

∴

S n9 = 20196

Sum of 9th Powers of First N Natural Numbers Formula Elements

Variables

Sum of 9th Powers of First N Natural Numbers

The Sum of 9th Powers of First N Natural Numbers is the summation of the 9th powers of the natural numbers starting from 1 to the nth natural number.

Symbol: S_n9

Measurement: NAUnit: Unitless

Note: Value should be greater than 0.

Formulas to find Sum of 9th Powers of First N Natural Numbers

Value of N

The Value of N is the total number of terms from the beginning of the series up to where the sum of series is calculating.

Symbol: n

Measurement: NAUnit: Unitless

Note: Value should be greater than 0.

Formulas to find Value of N

Other formulas in Sum of 4th Powers category

Go Sum of 4th Powers of First N Natural Numbers

S n4 = \frac{n \cdot (n + 1) \cdot (2 \cdot n + 1) \cdot (3 \cdot n^{2} + 3 \cdot n - 1)}{30}

Go Sum of 5th Powers of First N Natural Numbers

S n5 = \frac{n^{2} \cdot (2 \cdot n^{2} + 2 \cdot n - 1) \cdot {(n + 1)}^{2}}{12}

Go Sum of 6th Powers of First N Natural Numbers

S n6 = \frac{n \cdot (n + 1) \cdot (2 \cdot n + 1) \cdot (3 \cdot n^{4} + 6 \cdot n^{3} - 3 \cdot n + 1)}{42}

Go Sum of 7th Powers of First N Natural Numbers

S n7 = \frac{n^{2} \cdot (3 \cdot n^{4} + 6 \cdot n^{3} - n^{2} - 4 \cdot n + 2) \cdot {(n + 1)}^{2}}{24}

How to Evaluate Sum of 9th Powers of First N Natural Numbers?

Sum of 9th Powers of First N Natural Numbers evaluator uses Sum of 9th Powers of First N Natural Numbers = (Value of N^2*(Value of N^2+Value of N-1)*(2*Value of N^4+4*Value of N^3-Value of N^2-3*Value of N+3)*(Value of N+1)^2)/20 to evaluate the Sum of 9th Powers of First N Natural Numbers, The Sum of 9th Powers of First N Natural Numbers formula is defined as the summation of the 9th powers of the natural numbers starting from 1 to the nth natural number. Sum of 9th Powers of First N Natural Numbers is denoted by S_n9 symbol.

How to evaluate Sum of 9th Powers of First N Natural Numbers using this online evaluator? To use this online evaluator for Sum of 9th Powers of First N Natural Numbers, enter Value of N (n) and hit the calculate button.

FAQs on Sum of 9th Powers of First N Natural Numbers

What is the formula to find Sum of 9th Powers of First N Natural Numbers?

The formula of Sum of 9th Powers of First N Natural Numbers is expressed as Sum of 9th Powers of First N Natural Numbers = (Value of N^2*(Value of N^2+Value of N-1)*(2*Value of N^4+4*Value of N^3-Value of N^2-3*Value of N+3)*(Value of N+1)^2)/20. Here is an example- 20196 = (3^2*(3^2+3-1)*(2*3^4+4*3^3-3^2-3*3+3)*(3+1)^2)/20.

How to calculate Sum of 9th Powers of First N Natural Numbers?

With Value of N (n) we can find Sum of 9th Powers of First N Natural Numbers using the formula - Sum of 9th Powers of First N Natural Numbers = (Value of N^2*(Value of N^2+Value of N-1)*(2*Value of N^4+4*Value of N^3-Value of N^2-3*Value of N+3)*(Value of N+1)^2)/20.

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