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

∑(k2k=014-4)

Short Answer

Expert verified

The sum of this sequence is 955.

Step by step solution

01

Step 1. Write the given information. 

The sum of sequence:

∑(k2-4)k=014

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=114=∑(k2)k=114-∑(4)k=114

03

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

Using formula for summation is:

∑k2k=0n=12+22+...+n2∑k2k=0n=0+∑k2k=1n⇒∑k2k=0n=0+n(n+1)(2n+1)6

So,

∑k2k=014=14(14+1)((2×14)+1)6⇒∑k2k=014=60906⇒∑k2k=014=1015

04

Step 4. Use the formula for sum of sequences. 

Using formula for summation is:

∑ck=0n=c+c+...+c⇒∑ck=0n=cn,c-realnumberSo,∑4k=014=4×(14+1)⇒∑4k=014=4×15⇒∑4k=014=60


05

Step 5. Now, subtract Step 3  from Step 4.

From Step 3,

∑k2k=114=1015

From Step 4,

∑4k=114=60

So,

∑(k2-4)k=114=∑(k2)k=114-∑(4)k=114⇒∑(k2-4)k=114=1015-60⇒∑(k2-4)k=114=955

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