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In Problems 71-82, find the sum of each sequence.

∑(k2k=116+4)

Short Answer

Expert verified

The sum of this sequence is 1560.

Step by step solution

01

Step 1. Write the given information. 

The sum of sequences:

∑(k2+4)k=116=∑(k2)k=116+∑(4)k=116

02

Step 2. Use the properties of sequence.

Using the properties of sequence:

∑(ak+bk)k=1n=∑(ak)k=1n+∑(bk)k=1n

So,

∑(k2+4)k=116=∑(k2)k=116+∑(4)k=116

03

Step 3. Use the formula for sum of sequences of (n2 ) real numbers.

Using formula for summation is:

∑k2k=1n=12+22+...+n2⇒∑k2k=1n=n(n+1)(2n+1)6

So,

∑k2k=116=16(16+1)((2×16)+1)6⇒∑k2k=116=16×17×336⇒∑k2k=116=89766⇒∑k2k=116=1496

04

Step 4. Use the formula for sum of sequences. 

Using formula for summation is:
∑ck=1n=c+c+...+c⇒∑ck=1n=cn,c-realnumberSo,∑4k=116=4×16⇒∑4k=116=64

05

Step 5. Now, add the two sums from Step 3 and Step 4.

From Step 3,

∑k2k=116=1496

From Step 4,

∑4k=116=64

So,

∑(k2+4)k=116=∑(k2)k=116+∑(4)k=116⇒∑(k2+4)k=116=1496+64⇒∑(k2+4)k=116=1560

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