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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 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:

\(\begin{array}{l}{\binom {n}{k}}={\frac {n^{\underline {k}}}{k!}}={\frac {n(n-1)(n-2)\cdots (n-(k-1))}{k(k-1)(k-2)\cdots 1}}=\prod _{i=1}^{k}{\frac {n+1-i}{i}}\end{array} \)

3) Factorial formula:

\(\begin{array}{l}{\binom {n}{k}}={\frac {n!}{k!\,(n-k)!}}\quad {\text{for }}\ 0\leq k\leq n\end{array} \)

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 G.P. 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 G.P. 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 G.P. 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 expansions,

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: Coefficient of xr [0 ≤ r ≤ (n – 1)] in the expansion of

\(\begin{array}{l}{{(x+3)}^{n-1}}+{{(x+3)}^{n-2}}(x+2)+{{(x+3)}^{n-3}}{{(x+2)}^{2}}+…+{{(x+2)}^{n-1}}\end{array} \)

Solution:

\(\begin{array}{l}(x+3)^{n-1}+(x+3)^{n-2}(x+2)+(x+3)^{n-3}(x+2)^{2}+….+(x+2)^{n-1}\end{array} \)
\(\begin{array}{l}=\frac{(x+3)^{n}-(x+2)^{n}}{(x+3)-(x+2)} = (x+3)^{n}-(x+2)^{n}\end{array} \)
\(\begin{array}{l}(\because \frac{x^{n}-a^{n}}{x-a})=x^{n-1}+x^{n-2}a^{1}+x^{n-3}a^{2}+….+a^{n-1}\end{array} \)

Coefficient of xr in the given expression = Coefficient of xr in [(x + 3)n – (x + 2)n]

\(\begin{array}{l}= ^{n}C_{r}3^{n-r}\; -\; ^{n}C_{r}2^{n-r}\end{array} \)
\(\begin{array}{l}^{n}C_{r}(3^{n-r} – 2^{n-r})\end{array} \)

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

Solution:

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

Example 4:

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

Solution:

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

Binomial Coefficients with G.P. or A.P. Multiplied – Video Lesson

JEE Maths

Frequently Asked Questions

Q1

What do you mean by binomial coefficients?

Binomial coefficients are the positive integers which are the coefficients in the binomial theorem.

Q2

Give the binomial theorem formula.

The binomial theorem formula is given by (x+y)n = Σr=0n  nCr xn – r yr. n ∈ N, x,y,∈ R.

Q3

Give two applications of the binomial theorem.

The binomial theorem is used to find the expansions for the algebraic identities. It is also used in probability for binomial expansion.

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