Binomial Theorem

8. The general term of the expansion (1 + a)^m+n is

T_r+1 = ^m+nC_ra^r   [since, 1^m+n-r = 1]

At r = m we have,

T_m+1 = ^m+nC_ma^m

= $\frac{\left(m+n\right)!}{m!\left(m+n-m\right)!}$ (a)^m

= $\frac{\left(m+n\right)!}{m! n!}$ a^m - (1)

Similarly at r = n we have,

T_n+1 = ^m+nC_naⁿ

= $\frac{\left(m+n\right)!}{n!\left(m+n-n\right)!}$ (a)ⁿ

= $\frac{\left(m+n\right)!}{n! m!}$ aⁿ - (2)

Hence from (1) & (2),

Co-efficient of am = Co-efficient of aⁿ = $\frac{\left(m+n\right)!}{m!n!}$

6. Let (r + 1)th be the general term of ( $x^2- yx){}^{12}$

So, T_r-1 = ¹²C_r (x²)^12–r (–yx)^r

= (–1)^r12C_rx^24–2ry^rx^r

= (–1)^r12C_r $x^{24-2r+r}{y}^r$

= (-1)^r12C­_r $x^{24-r}{y}^r$

5. Let (r + 1)^th term be the general term of (x²–y)⁶.

So, T_r-1 = ⁶C_r (x²)^6-r (-y)^r

= (–1)^r .⁶C_r . $x^{12-r}$ . $y^r$

4. Let a⁵b⁷ occurs in (r + 1)^th term of [removed]a – 2b)¹².

Now, T_r+1 = ¹²C_ra^12-r (–2b)^r

= (–1)^r12C_ra^12–r . 2^r. b^r

Comparing indices of a and b in T_r-1 with a⁵ and b⁷ we get,  r = 7

So, co-efficient of a⁵b⁷ is (–1)⁷¹²C₇ 2⁷

2.

By binomial theorem,

(a + b)ⁿ = ⁿC₀ (a)ⁿ (b)⁰ + ⁿC₁ (a)^n–1 (b)¹ + …………. + ⁿC_r (a)^n–r (b)^r + …………… + ⁿC_n (a)^n–n (b)ⁿ

Where, b⁰ = 1 = a^n–n

So, (a + b)ⁿ = ⁿC_r (a)^n–r (b)^r

Putting a = 1 and b = 3 such that a + b = 4, we can rewrite the above equation as

(1 + 3)ⁿ = ⁿC_r (1)^n–r.3^r

=>4ⁿ = $\underset{r=0}{{\displaystyle ?}}.3r$ .ⁿC_r

Hence proved.