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Question: How do you find the partial sum of \(\sum {(2n - 1)} \) from n = 1 to 400?...

How do you find the partial sum of ∑(2n−1)\sum {(2n - 1)} from n = 1 to 400?

Explanation

Solution

According to the given question, we have to find the partial sum of ∑(2n−1)\sum {(2n - 1)} from n = 1 to 400. So, first of all we have to use the distributive property as mentioned below.
Distributive property: The distributive property is an algebraic property that is used to multiply a single value and two or more values within a set of parenthesis. The distributive property states that when factor is multiplied by sum/addition of two terms, it is essential to multiply each of the two numbers by the factor, and finally perform the addition operation. This property can be stated symbolically as,
⇒∑(A+B)=∑A+∑B\Rightarrow \sum {\left( {A + B} \right)} = \sum {A + } \sum BWhere, A, B are two different values.
Now, we have to use the first N integers given by gauss’ formula as mentioned below.
Gauss’ formula:
⇒∑n=1Nn=N(N+1)2.............................(A)\Rightarrow \sum\limits_{n = 1}^N n = \dfrac{{N\left( {N + 1} \right)}}{2}.............................(A)
While the sum of N times the unity is clearly N as mentioned below.
⇒∑n=1N1=N...................................(B)\Rightarrow \sum\limits_{n = 1}^N 1 = N...................................(B)

Complete step-by-step answer:
Step 1: First of all we have to use the distributive property of the sum as mentioned in the solution hint.
⇒∑n=1400(2n)−∑n=1400(1)\Rightarrow \sum\limits_{n = 1}^{400} {\left( {2n} \right)} - \sum\limits_{n = 1}^{400} {\left( 1 \right)}
Step 2: Now, we have to use the formula (A) and (B) as mentioned in the solution hint for the expression as obtained in the step 1.
⇒2×400(400+1)2−400\Rightarrow 2 \times \dfrac{{400\left( {400 + 1} \right)}}{2} - 400
Step 3: Now, we have to solve the expression obtained in the solution step 2.

⇒2×400×4012−400 ⇒2×200×401−400 ⇒16000  \Rightarrow 2 \times \dfrac{{400 \times 401}}{2} - 400 \\\ \Rightarrow 2 \times 200 \times 401 - 400 \\\ \Rightarrow 16000 \\\

Final solution: Hence, the partial sum of ∑(2n−1)\sum {(2n - 1)} from n = 1 to 400 is1600016000.

Note:
It is necessary to use the distributive property that is mentioned in the solution hint.
It is necessary to use the gauss’ formula to solve the expression obtained in the solution step 1.