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

∑(5k=120k+3)

Short Answer

Expert verified

The sum of this sequence is 1110.

Step by step solution

01

Step 1. Write the given information. 

The sum of sequence:

∑(5k+3)k=120

02

Step 2. Use the properties of sequence.

Using the properties of sequence:
∑(ak+bk)k=1n=∑(ak)k=1n+∑(bk)k=1n&∑(cak)k=1n=c∑akk=1n

So,

∑(5k+3)k=120=∑(5k)k=120+∑(3)k=120

&

∑(5k)k=120=5∑(k)k=120

03

Step 3. Use the formula for sum of sequences of n real numbers.

Using formula for summation is:

∑kk=1n=1+2+...+n⇒∑kk=1n=n(n+1)2

So,

5∑kk=120=5×20(20+1)2⇒5∑kk=1n=5×20(21)2⇒5∑kk=1n=5×210⇒5∑kk=1n=1050

04

Step 4. Use the formula for sum of sequence. 

Using formula for summation is:

∑ck=1n=c+c+...+c⇒∑ck=1n=cn,c-realnumberSo,∑3k=120=3×20⇒∑3k=120=60

05

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

From Step 3,

5∑kk=1n=1050

From Step 4,

∑3k=120=60

So,

localid="1646750229919" ∑(5k+3)k=120=∑(5k)k=120+∑(3)k=120⇒∑(5k+3)k=120=1050+60⇒∑(5k+3)k=120=1110

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