# Binomial Coefficients with G.P. or A.P. Multiplied

The positive integers that are present as coefficients in the binomial expansion are termed binomial coefficients. It was introduced in the year 1826. The Indian Mathematician named Bhaskaracharya was the person who gave the description of the coefficients in the binomial expansion. The binomial coefficients can be computed using the following formulae:

1] Recursive formula: nCk = n-1Ck-1 + n-1Ck for all integers n.

2] Multiplicative formula: ${\binom {n}{k}}={\frac {n^{\underline {k}}}{k!}}={\frac {n(n-1)(n-2)\cdots (n-(k-1))}{k(k-1)(k-2)\cdots 1}}=\prod _{i=1}^{k}{\frac {n+1-i}{i}}$

3] Factorial formula: ${\binom {n}{k}}={\frac {n!}{k!\,(n-k)!}}\quad {\text{for }}\ 0\leq k\leq n$

4] Pascal’s triangle consists of the value of binomial coefficients.

Binomial Coefficients with G.P. or A.P. Multiplied

Basic idea:

Consider (1 + x)n = nC0 + nC1 x + nC2 x2 + nC3 x3 + …….

Where x1, x2, x3 ….. are in GP multiplied with nC0, nC1, ….

• Start from nC0.
• x is a variable multiplied by nC1

Example 1: nC0 + nC1 3 + nC2 32 + nC3 33 + …..

Solution:

By comparing the given problem to (1 + x)n = nC0 + nC1 x + nC2 x2 + nC3 x3 + …….,

the answer is (1 + 3)n.

Example 2: nC0 + nC1 21 + nC2 22 + ……

Solution:

It is a GP with a common ratio 2 = x.

nC0 + nC1 21 + nC2 22 + ……

= (1 + 2)n

= 3n

Example 3: nC0 + nC1 42 + nC2 44 + ……

Solution:

It is a GP with a common ratio 42 = x.

nC0 + nC1 42 + nC2 44 + ……

(1 + 42)n = 17n

The case when AP is multiplied by the binomial coefficient.

nC0 a + nC1 (a + d) + nC2 (a + 2d) + ……… + nCn (a + nd)

= {[a (a + nd)] / 2} {nC0 + nC1 + ………. + nCn}

= {[a (a + nd)] / 2} (2n)

= a + (a + nd) 2n-1

Proof:

S = nC0 a + nC1 (a + d) + nC2 (a + 2d) + ….. + nCn (a + nd)

S = nCn (a + nd) + nCn-1 [(a + (n – 1) d] + ……. + nC0 a

On adding the above two expansion,

2S = nC0 [a + (a + nd)] + nC1 [a + (a + nd)] + nCn-2 [a + (a + nd)] + …… + nCn [a + (a + nd)]

= [a + (a + nd)] [nC0 + nC1 + nC2 + ……. + nCn]

S = {[a (a + nd)] / 2} (2n)

Example 1: nC0 + 5 . nC1 + 7 . nC2 + …….. + (2n + 3) . nCn = ?

Solution:

{[3 + (2n + 3)] / 2} {2n-1}

= (2n + 6) 2n-1

= (n + 3) 2n

Example 2: Coefficients of ${{x}^{r}}[0\le r\le (n-1)] \text \ in \ the \ expansion \ of \ {{(x+3)}^{n-1}}+{{(x+3)}^{n-2}}(x+2)+{{(x+3)}^{n-3}}{{(x+2)}^{2}}+…+{{(x+2)}^{n-1}}$

Solution:

$(x+3)^{n-1}+(x+3)^{n-2}(x+2)+(x+3)^{n-3}(x+2)^{2}+….+(x+2)^{n-1}$ $=\frac{(x+3)^{n}-(x+2)^{n}}{(x+3)-(x+2)} = (x+3)^{n}-(x+2)^{n}$ $(\because \frac{x^{n}-a^{n}}{x-a})=x^{n-1}+x^{n-2}a^{1}+x^{n-3}a^{2}+….+a^{n-1}$ $Coefficient\; of\; {x}^{r}\: in\: the \: given \: expression = Coefficient\; of\; {x}^{r}\; in\; [(x+3)^{n}-(x+2)^{n}]$ $= ^{n}C_{r}3^{n-r}\; -\; ^{n}C_{r}2^{n-r}$ $^{n}C_{r}(3^{n-r} – 2^{n-r})$

Example 3: The sum of the last eight coefficients in the expansion of (1 + x)15 is

Solution:

$^{15}{{C}_{0}}+{{\,}^{15}}{{C}_{1}}+…+{{\,}^{15}}{{C}_{15}}={{2}^{15}}\\ \Rightarrow \,\,\,2{{(}^{15}}{{C}_{8}}+{{\,}^{15}}{{C}_{9}}+…+{{\,}^{15}}{{C}_{15}})={{2}^{15}} (\because \,{{\,}^{n}}{{C}_{r}}={{\,}^{n}}{{C}_{n-r}})\\ ^{15}{{C}_{8}}+{{\,}^{15}}{{C}_{9}}+…+{{\,}^{15}}{{C}_{15}}\\ ={{2}^{14}}$

Example 4: If ${{a}_{k}}=\frac{1}{k(k+1)}, \text \ for \ k=1,\,2,\,3,\,4,…..,\,n, \ then {{\left( \sum\limits_{k=1}^{n}{{{a}_{k}}} \right)}^{2}}=$

Solution:

$\sum\limits_{k=1}^{n}{{{a}_{k}}}=\sum\limits_{k=1}^{n}{\frac{1}{k\,(k+1)}}\\ =\left( 1-\frac{1}{2} \right)+\left( \frac{1}{2}-\frac{1}{3} \right)+…+\left( \frac{1}{n}-\frac{1}{n+1} \right)\\ = 1-\frac{1}{n+1}=\frac{n}{n+1}{{\left( \sum\limits_{k=1}^{n}{{{a}_{k}}} \right)}^{2}}\\ ={{\left( \frac{n}{n+1} \right)}^{2}}$