Question

✬ HeY There!

In the expansion of (1 + a)ᵐ ⁺ ⁿ prove that coefficients of aᵐ and aⁿ are equal.​

Answer 1

★ Concept ::

Here the concept of Binomial Expansion has been used. We see that we are given an equation and we have to prove that if the coefficient of given terms are equal in the expansion of given equation or not. So firstly, we can find the coefficient of aᵐ and aⁿ and then equate them to prove they are equal.

Let's do it !!

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★ Formula Used :-

$\\\;\boxed{\sf{\pink{T_{r\:+\:1}\;=\;\bf{^{p}C_{r}\:b^{p\:-\:1}\:c^{r}}}}}$

$\\\;\boxed{\sf{\pink{^{p}C_{r}\;=\;\bf{\dfrac{p!}{r!(p\:-\:r)!}}}}}$

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★ Solution :-

Given,

✒ (1 + a)ᵐ ⁺ ⁿ

We know that,

$\\\;\sf{\rightarrow\;\;T_{r\:+\:1}\;=\;\bf{^{p}C_{r}\:b^{p\:-\:1}\:c^{r}}}$

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~ For the coefficient of aᵐ ::

Let us assume that aᵐ comes in (r + 1)th which is in the expansion of (1 + a)ᵐ ⁺ ⁿ. So we have the formula as,

$\\\;\sf{\rightarrow\;\;T_{r\:+\:1}\;=\;\bf{^{p}C_{r}\:b^{p\:-\:1}\:c^{r}}}$

• Here b = 1 and c = a

• p = m + n

By applying values, we get

$\\\;\sf{\rightarrow\;\;T_{r\:+\:1}\;=\;\bf{^{m\:+\:n}C_{r}\:(1)^{m\:+\:n\:-\:1}\:(a)^{r}}}$

Since, when 1 is raised to any coefficient the result is 1 only. So,

$\\\;\sf{\rightarrow\;\;T_{r\:+\:1}\;=\;\bf{^{m\:+\:n}C_{r}\:(1)\:(a)^{r}}}$

$\\\;\bf{\rightarrow\;\;\red{T_{r\:+\:1}\;=\;\bf{^{m\:+\:n}C_{r}\:a^{r}}}}$

Now let's compare the coefficient of a in aᵐ and T_(r + 1) , we get

✒ aᵐ = a^{r}

✒ r = m

By the formula of Combinations, we know that

$\\\;\sf{\Longrightarrow\;\;^{p}C_{r}\;=\;\bf{\dfrac{p!}{r!(p\:-\:r)!}}}$

• Here p = m + n

• r = m

By applying the values, we get

$\\\;\sf{\Longrightarrow\;\;^{m\:+\:n}C_{m}\;=\;\bf{\dfrac{(m\:+\:n)!}{m!(m\:+\:n\:-\:m)!}}}$

$\\\;\sf{\Longrightarrow\;\;^{m\:+\:n}C_{m}\;=\;\bf{\dfrac{(m\:+\:n)!}{m!(n)!}}}$

$\\\;\sf{\Longrightarrow\;\;\blue{ ^{m\:+\:n}C_{m}\;=\;\bf{\dfrac{(m\:+\:n)!}{m!\:n!}}}}$

Let this be equation i)

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~ For the coefficient of aⁿ ::

Let us assume that aⁿ comes in the (k + 1)th term which comes in the expansion of (1 + a)ᵐ ⁺ ⁿ. So by the formula, we get

$\\\;\sf{\rightarrow\;\;T_{k\:+\:1}\;=\;\bf{^{p}C_{k}\:b^{p\:-\:1}\:c^{k}}}$

• Here p = m + n

• Here b = 1 and c = a

By applying values, we get

$\\\;\sf{\rightarrow\;\;T_{k\:+\:1}\;=\;\bf{^{m\:+\:n}C_{k}\:(1)^{m\:+\:n\:-\:1}\:(a)^{k}}}$

Since, when 1 is raised to any exponent, then the result is 1 only. So,

$\\\;\sf{\rightarrow\;\;T_{k\:+\:1}\;=\;\bf{^{m\:+\:n}C_{k}\:(1)\:(a)^{k}}}$

$\\\;\sf{\rightarrow\;\;\orange{T_{k\:+\:1}\;=\;\bf{^{m\:+\:n}C_{k}\:a^{k}}}}$

On comparing the coefficient of a in aⁿ and T_(k + 1), we get

✒ aⁿ = a^{k}

✒ n = k

By using the formula of Combinations, we get

$\\\;\sf{\Longrightarrow\;\;^{p}C_{k}\;=\;\bf{\dfrac{p!}{k!(p\:-\:k)!}}}$

• Here p = m + n

• Here k = n

By applying values, we get

$\\\;\sf{\Longrightarrow\;\;^{m\:+\:n}C_{n}\;=\;\bf{\dfrac{(m\:+\:n)!}{n!(m\:+\:n\:-\:n)!}}}$

$\\\;\sf{\Longrightarrow\;\;^{m\:+\:n}C_{n}\;=\;\bf{\dfrac{(m\:+\:n)!}{n!(m)!}}}$

$\\\;\sf{\Longrightarrow\;\;^{m\:+\:n}C_{n}\;=\;\bf{\dfrac{(m\:+\:n)!}{n!\:m!}}}$

$\\\;\bf{\Longrightarrow\;\;\green{^{m\:+\:n}C_{n}\;=\;\bf{\dfrac{(m\:+\:n)!}{m!\:n!}}}}$

Let this be equation ii).

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~ For final proving ::

On comparing equation one, we get

$\\\;\bf{\Longrightarrow\;\;\dfrac{(m\:+\:n)!}{m!\:n!}\;=\;\dfrac{(m\:+\:n)!}{m!\:n!}}$

From this, we get

$\\\;\bf{\Longrightarrow\;\;\purple{Coefficient\;of\;a^{m}\;=\;Coefficient\;of\;a^{n}}}$

Hence, proved.

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Answer 2

It is known that (r+1) the term (T r+1 ) in the binomial expansion of (a+b) n is given by

T r+1 = n C ra n−1b r

Assuming that a moccurs in the (r+1) the term of the expansion (1+a) m+n we obtain

T r+1 = m+n C r

(1) m+n−r

(a)

r

=

m+n

C

r

a

r

Comparing the indices of a in a

m

and in T

r+1

we obtain r=m

Therefore the coefficient of a

m

is

m+n

C

m

=

m!(m+n−m)!

(m+n)!

=

m!n!

(m+n)!

....(1)

Assuming that a

n

occurs in the (k+1)th term of expansion (1+a)

m+n

, we obtain

T

k+1

=

m+n

C

k

(1)

m+n−k

(a)

k

=

m+n

C

k

a

k

Comparing the indices of a in a

n

and in T

k+1

we obtain k=n

Therefore the coefficient of a

n

is

m+n

C

n

=

n!(m+n−n)!

(m+n)!

=

n!m!

(m+n)!

....(2)

Thus from 1 and 2 it can be observed that the coefficients of a

m

and a

n

in the expansion of (1+a)

m+nare equal.