Lesson The Pascal's triangle

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<H2>The Pascal's triangle</H2>

This lesson is about the Pascal's triangle. &nbsp;You should know what the binomial coefficients {{{C[n]^m}}} are. 
You can find all the necessary information in the lesson &nbsp;<A HREF=http://www.algebra.com/algebra/homework/Permutations/Binomial-Theorem.lesson>Binomial Theorem, Binomial Formula, Binomial Coefficients and Binomial Expansion</A>&nbsp; under the current topic in this site.

<BLOCKQUOTE><B>Definition</B>. &nbsp;The Pascal's triangle is the triangle of numbers formed by the binomial coefficients &nbsp;{{{C[n]^m}}}&nbsp; in a way shown in the &nbsp;<B>Figure 1</B>&nbsp; and the &nbsp;<B>Figure 2</B>&nbsp; below.

<TABLE> 
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&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;{{{C[0]^0}}}

&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;{{{C[3]^0}}} &nbsp;&nbsp;&nbsp;&nbsp;{{{C[3]^1}}} &nbsp;&nbsp;&nbsp;&nbsp;{{{C[3]^2}}} &nbsp;&nbsp;&nbsp;&nbsp;{{{C[3]^3}}}

&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;{{{C[4]^0}}} &nbsp;&nbsp;&nbsp;&nbsp;{{{C[4]^1}}} &nbsp;&nbsp;&nbsp;&nbsp;{{{C[4]^2}}} &nbsp;&nbsp;&nbsp;&nbsp;{{{C[4]^3}}} &nbsp;&nbsp;&nbsp;&nbsp;{{{C[4]^4}}}

&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;{{{C[5]^0}}} &nbsp;&nbsp;&nbsp;&nbsp;{{{C[5]^1}}} &nbsp;&nbsp;&nbsp;&nbsp;{{{C[5]^2}}} &nbsp;&nbsp;&nbsp;&nbsp;{{{C[5]^3}}} &nbsp;&nbsp;&nbsp;&nbsp;{{{C[5]^4}}} &nbsp;&nbsp;&nbsp;&nbsp;{{{C[5]^5}}}

&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;{{{C[n]^0}}} &nbsp;&nbsp;&nbsp;&nbsp;{{{C[n]^1}}} &nbsp;&nbsp;&nbsp;&nbsp;{{{C[n]^2}}} &nbsp;&nbsp;. . . . .&nbsp;&nbsp;&nbsp;{{{C[n]^(n-2)}}} &nbsp;{{{C[n]^(n-1)}}} &nbsp;{{{C[n]^n}}}

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

<B>Figure 1</B>. The Pascal's triangle in terms of binomial coefficients {{{C[n]^m}}}&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;
 </TD>
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&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;{{{1}}}

&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;{{{1}}} &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;{{{3}}} &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;{{{3}}} &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;{{{1}}}

&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;{{{1}}} &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;{{{4}}} &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;{{{6}}} &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;{{{4}}} &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;{{{1}}}

&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;{{{1}}} &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;{{{5}}} &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;{{{10}}} &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;{{{10}}} &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;{{{5}}} &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;{{{1}}}

&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;<B>Figure 2</B>. The Pascal's triangle in numerical terms
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The upper row contains only one number &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;{{{C[0]^0}}} = {{{1}}}. &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;(By the agreement, this row has the number 0, "zero".)

The next row contains two numbers &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;{{{C[1]^0}}} = {{{1}}}&nbsp; and &nbsp;{{{C[1]^1}}} = {{{1}}}.&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;(This row has the number 1.)

The row number 2 contains three numbers &nbsp;&nbsp;{{{C[2]^0}}} = {{{1}}}, &nbsp;{{{C[2]^1}}} = {{{2}}} &nbsp;and &nbsp;{{{C[2]^2}}} = {{{1}}}.

The row number 3 contains four numbers &nbsp;&nbsp;&nbsp;&nbsp;{{{C[3]^0}}} = {{{1}}}, &nbsp;{{{C[3]^1}}} = {{{3}}}, &nbsp;{{{C[3]^2}}} = {{{3}}}  &nbsp;and &nbsp;{{{C[3]^3}}} = {{{1}}}.

The row 4 contains five numbers &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;{{{C[4]^0}}} = {{{1}}}, &nbsp;{{{C[4]^1}}} = {{{4}}}, &nbsp;{{{C[4]^2}}} = {{{6}}},  &nbsp;{{{C[4]^3}}} = {{{4}}} &nbsp;and &nbsp;{{{C[4]^4}}} = {{{1}}}.

The row 5 contains six numbers &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;{{{C[5]^0}}} = {{{1}}}, &nbsp;{{{C[5]^1}}} = {{{5}}}, &nbsp;{{{C[5]^2}}} = {{{10}}},  &nbsp;{{{C[5]^3}}} = {{{10}}}, &nbsp;{{{C[5]^4}}} = {{{5}}} &nbsp;and&nbsp; {{{C[5]^5}}} = {{{1}}}.

And so on ...

The &nbsp;<B>n</B>-th &nbsp;row contains numbers &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;{{{C[n]^0}}} = {{{1}}}, &nbsp;{{{C[n]^1}}} = {{{n}}}, &nbsp;{{{C[n]^2}}} = {{{(n*(n-1))/2}}},  &nbsp;{{{C[n]^3}}} = {{{(n*(n-1)*(n-2))/(1*2*3)}}}, . . . , &nbsp;{{{C[n]^m}}} = {{{n!/((n-m)!*m!))}}} = {{{(n*(n-1)*(n-2)* ellipsis *(n-m+1))/(1*2*3* ellipsis *m)}}}, . . . ,

&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;{{{C[n]^(n-2)}}}={{{(n*(n-1))/2}}}, &nbsp;{{{C[n]^(n-1)}}} = {{{n}}} &nbsp;and &nbsp;{{{C[n]^n}}} = {{{1}}}.</BLOCKQUOTE>

<H3>The major property of the Pascal's triangle</H3><TABLE> 
  <TR>
  <TD>
In the Pascal triangle, the very first and the very last&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;
number in each row is equal to 1.

<B>Starting from the row number 2, each number between 
the very first and very last is equal to the sum of two 
its closest neighbors in the previous row</B>.

In mathematical terms, this means that

{{{C[n]^k}}} + {{{C[n]^(k+1)}}} = {{{C[n+1]^(k+1)}}}

for any positive integer number &nbsp;<B>n</B>&nbsp; and all integer 
numbers k = 0, 1, 2, . . . , <B>n</B>.

This property is illustrated in the &nbsp;<B>Figure 3</B>.

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&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;<B>Figure 3</B>. The property of the Pascal's triangle 
  </TD>
  </TR>
</TABLE>

<B>Proof</B>

Actually, &nbsp;this identity was just proved in the lesson &nbsp;<A HREF=http://www.algebra.com/algebra/homework/Permutations/Binomial-Theorem.lesson>Binomial Theorem, Binomial Formula, Binomial Coefficients and Binomial Expansion</A>&nbsp; under the current topic in this site.
For your convenience, &nbsp;here we repeat the proof again.

We have

{{{C[n]^k}}} + {{{C[n]^(k+1)}}} = {{{(n*(n-1)* ellipsis *(n-k+1))/(1*2* ellipsis *k)}}} + {{{(n*(n-1)* ellipsis *(n-k+1)*(n-k))/(1*2* ellipsis *k*(k+1))}}} = &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;(now express fractions as equivalent fractions with the common denominator &nbsp;{{{1*2* ellipsis *k*(k+1)}}})

{{{(n*(n-1)* ellipsis *(n-k+1)*(k+1))/(1*2* ellipsis *k*(k+1))}}} + {{{(n*(n-1)* ellipsis *(n-k+1)*(n-k))/(1*2* ellipsis *k*(k+1))}}} = &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;(now take the common factor &nbsp;{{{(n*(n-1)* ellipsis *(n-k+1))/(1*2* ellipsis *k*(k+1))}}} out of brackets)

{{{(n*(n-1)* ellipsis *(n-k+1))/(1*2* ellipsis *k*(k+1))}}} * {{{((k+1) + (n-k))}}} = &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;(now rewrite the expression in the very right brackets)

{{{(n*(n-1)* ellipsis *(n-k+1))/(1*2* ellipsis *k*(k+1))}}} * {{{(n+1-(k+1)+1)}}} =

{{{(n*(n-1)* ellipsis *(n-k+1)*(n+1-(k+1)+1))/(1*2* ellipsis *k*(k+1))}}}  = {{{C[n+1]^(k+1)}}}.

We got the required identity. &nbsp;The proof is completed.

<H3>Other properties of the Pascal's triangle</H3>

<BLOCKQUOTE><B>1. &nbsp;In the Pascals triangle the sum of all numbers in the n-th row is equal to {{{2^n}}} </B>.

<B>2. &nbsp;In the Pascals triangle the alternate sum of all numbers in the n-th row is equal to 0</B>.</BLOCKQUOTE>

Let us prove the first of the two properties.
The sum of all numbers of the &nbsp;<B>n</B>-th &nbsp;row of the Pascal's triangle is equal to

{{{C[n]^0}}} + {{{C[n]^1}}} + {{{C[n]^2}}} + . . . + {{{C[n]^k}}} + . . . + {{{C[n]^n}}}.

We have to prove that

{{{C[n]^0}}} + {{{C[n]^1}}} + {{{C[n]^2}}} + . . . + {{{C[n]^k}}} + . . . + {{{C[n]^n}}} = {{{2^n}}}.

Use the binomial expansion

{{{(1 + x)^n}}} = &nbsp;&nbsp;=&nbsp;&nbsp; {{{ sum ( C[n]^i*x^i , i=0, n ) }}}

(see the lesson &nbsp;<A HREF=http://www.algebra.com/algebra/homework/Permutations/Binomial-Theorem.lesson>Binomial Theorem, Binomial Formula, Binomial Coefficients and Binomial Expansion</A>&nbsp; under the current topic in this site). &nbsp;Substitute &nbsp;{{{x}}} = {{{1}}} &nbsp;to the last

formula, and you will get the required identity.

This is how it looks for the low order binomial coefficients and the first rows of the Pascal's triangle:

<B>n</B> = 1, &nbsp;{{{C[1]^0}}} = 1, &nbsp;{{{C[1]^1}}} = 1, &nbsp;{{{C[1]^0}}} + {{{C[1]^1}}} = 1 + 1 = 2;

<B>n</B> = 2, &nbsp;{{{C[2]^0}}} = 1, &nbsp;{{{C[2]^1}}} = 2, &nbsp;{{{C[2]^2}}} = 1, &nbsp;{{{C[2]^0}}} + {{{C[2]^1}}} + {{{C[2]^2}}} = 1 + 2 + 1 = 4;

<B>n</B> = 3, &nbsp;{{{C[3]^0}}} = 1, &nbsp;{{{C[3]^1}}} = 3, &nbsp;{{{C[3]^2}}} = 3, &nbsp;{{{C[3]^3}}} = 1, &nbsp;{{{C[3]^0}}} + {{{C[3]^1}}} + {{{C[3]^2}}} + {{{C[3]^3}}} = 1 + 3 + 3 + 1 = 8.

The proof of the second property is also straightforward. 
The alternate sum of all numbers of the &nbsp;<B>n</B>-th &nbsp;row of the Pascal's triangle is equal to

{{{C[n]^0}}} - {{{C[n]^1}}} + {{{C[n]^2}}} + . . . + {{{(-1)^k*C[n]^k}}} + . . . + {{{(-1)^n*C[n]^n}}}.

We have to prove that

{{{C[n]^0}}} - {{{C[n]^1}}} + {{{C[n]^2}}} + . . . + {{{(-1)^k*C[n]^k}}} + . . . + {{{(-1)^n*C[n]^n}}} = {{{0}}}.

Use the same binomial expansion

{{{(1 + x)^n}}} = &nbsp;&nbsp;=&nbsp;&nbsp; {{{ sum ( C[n]^i*x^i , i=0, n ) }}}.

Substitute &nbsp;{{{x}}} = {{{-1}}} &nbsp;to the last formula, and you will get the required identity.

This is how it looks for the low order binomial coefficients and the first rows of the Pascal's triangle:

<B>n</B> = 1, &nbsp;{{{C[1]^0}}} = 1, &nbsp;{{{C[1]^1}}} = 1, &nbsp;{{{C[1]^0}}} - {{{C[1]^1}}} = 1 - 1 = 0;

<B>n</B> = 2, &nbsp;{{{C[2]^0}}} = 1, &nbsp;{{{C[2]^1}}} = 2, &nbsp;{{{C[2]^2}}} = 1, &nbsp;{{{C[2]^0}}} - {{{C[2]^1}}} + {{{C[2]^2}}} = 1 - 2 + 1 = 0;

<B>n</B> = 3, &nbsp;{{{C[3]^0}}} = 1, &nbsp;{{{C[3]^1}}} = 3, &nbsp;{{{C[3]^2}}} = 3, &nbsp;{{{C[3]^3}}} = 1, &nbsp;{{{C[3]^0}}} - {{{C[3]^1}}} + {{{C[3]^2}}} - {{{C[3]^3}}} = 1 - 3 + 3 - 1 = 0.

There are other identities for the binomial coefficients and the Pascal's triangle elements. 
For example,

{{{C[n]^0}}} + {{{2*C[n]^1}}} + {{{2^2*C[n]^2}}} + . . . + {{{2^k*C[n]^k}}} + . . . + {{{2^n*C[n]^n}}} = {{{3^n}}},

{{{C[n]^0}}} + {{{3*C[n]^1}}} + {{{3^2*C[n]^2}}} + . . . + {{{3^k*C[n]^k}}} + . . . + {{{3^n*C[n]^n}}} = {{{4^n}}},

{{{C[n]^0}}} + {{{(1/2)*C[n]^1}}} + {{{(1/2^2)*C[n]^2}}} + . . . + {{{(1/2^k)*C[n]^k}}} + . . . + {{{(1/2^n)*C[n]^n}}} = {{{(3/2)^n}}},

{{{C[n]^0}}} - {{{(1/2)*C[n]^1}}} + {{{(1/2^2)*C[n]^2}}} + . . . + {{{(((-1)^k)/2^k)*C[n]^k}}} + . . . + {{{(((-1)^n)/2^n)*C[n]^n}}} = {{{1/2^n}}}

for any positive integer number &nbsp;<B>n</B>.

You can prove these and similar identities yourself. &nbsp;Simply use the binomial expansion and make an appropriate substitution.

My lessons on Binomial Theorem, Binomial Formula, Binomial Coefficients and Binomial Expansion in this site are

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF =http://www.algebra.com/algebra/homework/Permutations/Binomial-Theorem.lesson>Binomial Theorem, Binomial Formula, Binomial Coefficients and Binomial Expansion</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF =http://www.algebra.com/algebra/homework/Permutations/Remarkable-identities-for-Binomial-Coefficients.lesson>Remarkable identities for Binomial Coefficients</A>

&nbsp;&nbsp;&nbsp;&nbsp;- The Pascal's triangle&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;(this lesson)

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/Permutations/Solved-problems-on-binomial-coefficients.lesson>Solved problems on binomial coefficients</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/Permutations/Solving-equations-that-include-binom-coefts-and-numbers-of-permutations.lesson>Solving equations that include binomial coefficients and numbers of permutations</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/Permutations/Math-circle-level-problem-on-binomial-coefficients.lesson>Math circle level problem on binomial coefficients</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF =https://www.algebra.com/algebra/homework/Permutations/OVERVIEW-of-lessons-on-Binomial-Expansion-Binomial-coefficients-and-Pascal%27s-triangle.lesson>OVERVIEW of lessons on Binomial Expansion, Binomial coefficients and the Pascal's triangle</A>

Use this file/link &nbsp;<A HREF=https://www.algebra.com/algebra/homework/complex/ALGEBRA-II-YOUR-ONLINE-TEXTBOOK.lesson>ALGEBRA-II - YOUR ONLINE TEXTBOOK</A>&nbsp; to navigate over all topics and lessons of the online textbook &nbsp;ALGEBRA-II.