Finite Math Examples

Expand Using the Binomial Theorem (2k+1)^3
(2k+1)3(2k+1)3
Step 1
Use the binomial expansion theorem to find each term. The binomial theorem states (a+b)n=nk=0nCk(an-kbk)(a+b)n=nk=0nCk(ankbk).
3k=03!(3-k)!k!(2k)3-k(1)k3k=03!(3k)!k!(2k)3k(1)k
Step 2
Expand the summation.
3!(3-0)!0!(2k)3-0(1)0+3!(3-1)!1!(2k)3-1(1)1+3!(3-2)!2!(2k)3-2(1)2+3!(3-3)!3!(2k)3-3(1)33!(30)!0!(2k)30(1)0+3!(31)!1!(2k)31(1)1+3!(32)!2!(2k)32(1)2+3!(33)!3!(2k)33(1)3
Step 3
Simplify the exponents for each term of the expansion.
1(2k)3(1)0+3(2k)2(1)1+3(2k)1(1)2+1(2k)0(1)31(2k)3(1)0+3(2k)2(1)1+3(2k)1(1)2+1(2k)0(1)3
Step 4
Simplify each term.
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Step 4.1
Multiply 11 by (1)0(1)0 by adding the exponents.
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Step 4.1.1
Move (1)0(1)0.
(1)01(2k)3+3(2k)2(1)1+3(2k)1(1)2+1(2k)0(1)3(1)01(2k)3+3(2k)2(1)1+3(2k)1(1)2+1(2k)0(1)3
Step 4.1.2
Multiply (1)0(1)0 by 11.
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Step 4.1.2.1
Raise 11 to the power of 11.
(1)011(2k)3+3(2k)2(1)1+3(2k)1(1)2+1(2k)0(1)3(1)011(2k)3+3(2k)2(1)1+3(2k)1(1)2+1(2k)0(1)3
Step 4.1.2.2
Use the power rule aman=am+naman=am+n to combine exponents.
10+1(2k)3+3(2k)2(1)1+3(2k)1(1)2+1(2k)0(1)310+1(2k)3+3(2k)2(1)1+3(2k)1(1)2+1(2k)0(1)3
10+1(2k)3+3(2k)2(1)1+3(2k)1(1)2+1(2k)0(1)310+1(2k)3+3(2k)2(1)1+3(2k)1(1)2+1(2k)0(1)3
Step 4.1.3
Add 00 and 11.
11(2k)3+3(2k)2(1)1+3(2k)1(1)2+1(2k)0(1)311(2k)3+3(2k)2(1)1+3(2k)1(1)2+1(2k)0(1)3
11(2k)3+3(2k)2(1)1+3(2k)1(1)2+1(2k)0(1)311(2k)3+3(2k)2(1)1+3(2k)1(1)2+1(2k)0(1)3
Step 4.2
Simplify 11(2k)311(2k)3.
(2k)3+3(2k)2(1)1+3(2k)1(1)2+1(2k)0(1)3(2k)3+3(2k)2(1)1+3(2k)1(1)2+1(2k)0(1)3
Step 4.3
Apply the product rule to 2k2k.
23k3+3(2k)2(1)1+3(2k)1(1)2+1(2k)0(1)323k3+3(2k)2(1)1+3(2k)1(1)2+1(2k)0(1)3
Step 4.4
Raise 22 to the power of 33.
8k3+3(2k)2(1)1+3(2k)1(1)2+1(2k)0(1)38k3+3(2k)2(1)1+3(2k)1(1)2+1(2k)0(1)3
Step 4.5
Apply the product rule to 2k2k.
8k3+3(22k2)(1)1+3(2k)1(1)2+1(2k)0(1)38k3+3(22k2)(1)1+3(2k)1(1)2+1(2k)0(1)3
Step 4.6
Raise 22 to the power of 22.
8k3+3(4k2)(1)1+3(2k)1(1)2+1(2k)0(1)38k3+3(4k2)(1)1+3(2k)1(1)2+1(2k)0(1)3
Step 4.7
Multiply 44 by 33.
8k3+12k2(1)1+3(2k)1(1)2+1(2k)0(1)38k3+12k2(1)1+3(2k)1(1)2+1(2k)0(1)3
Step 4.8
Evaluate the exponent.
8k3+12k21+3(2k)1(1)2+1(2k)0(1)38k3+12k21+3(2k)1(1)2+1(2k)0(1)3
Step 4.9
Multiply 1212 by 11.
8k3+12k2+3(2k)1(1)2+1(2k)0(1)38k3+12k2+3(2k)1(1)2+1(2k)0(1)3
Step 4.10
Simplify.
8k3+12k2+3(2k)(1)2+1(2k)0(1)3
Step 4.11
Multiply 2 by 3.
8k3+12k2+6k(1)2+1(2k)0(1)3
Step 4.12
One to any power is one.
8k3+12k2+6k1+1(2k)0(1)3
Step 4.13
Multiply 6 by 1.
8k3+12k2+6k+1(2k)0(1)3
Step 4.14
Multiply 1 by (1)3 by adding the exponents.
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Step 4.14.1
Move (1)3.
8k3+12k2+6k+(1)31(2k)0
Step 4.14.2
Multiply (1)3 by 1.
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Step 4.14.2.1
Raise 1 to the power of 1.
8k3+12k2+6k+(1)311(2k)0
Step 4.14.2.2
Use the power rule aman=am+n to combine exponents.
8k3+12k2+6k+13+1(2k)0
8k3+12k2+6k+13+1(2k)0
Step 4.14.3
Add 3 and 1.
8k3+12k2+6k+14(2k)0
8k3+12k2+6k+14(2k)0
Step 4.15
Simplify 14(2k)0.
8k3+12k2+6k+14
Step 4.16
One to any power is one.
8k3+12k2+6k+1
8k3+12k2+6k+1
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