Question

binomial theorem question​

Answer 1

here's the answer :)

hope it's a little helpful and understandable since I used factors here.

Answer 2

★Given :

• The coefficients of 5th,6th&7th terms in the expansion of (1+x)ⁿ are in AP.

★To find :

• Value of n.

★Solution :

We know,

$\displaystyle \sf (1+x)^n\longrightarrow \ ^nC_0 + ^nC_1x+^nC_2x^2...^nC_nx^n$

According to question,

$\sf T_5,T_6,T_7$ are in AP.Then,

$\rightarrow \sf T_6-T_5 = T_7-T_6 \\\\\rightarrow \sf T_6 + T_6 = T_7 + T_5 \\\\\rightarrow \sf 2T_6= T_7+T_5\longrightarrow (1)$

And we know,

• $\sf T_5 = \ ^nC_4$
• $\sf T_6 = \ ^nC_5$
• $\sf T_7 = \ ^nC_6$

Substituting these in equation 1,

$\rightarrow \sf 2\ ^nC_5 = \ ^nC_6+\ ^nC_4$

$\displaystyle \sf \rightarrow 2\ \frac{n\ !}{(n-5)!5!} = \frac{n\ !}{(n-4)!4!} +\frac{n\ !}{(n-6)!6!}$

$\displaystyle \sf \rightarrow \frac{2}{(n-5)5} = \frac{1}{(n-4)(n-5)} + \frac{1}{6. 5}$

$\displaystyle \sf \rightarrow \frac{2}{\cancel{(n-5)}\cancel{5}} = \frac{30+(n-4)(n-5)}{6.\cancel{5}(n-4)\cancel{(n-5)}}$

$\displaystyle \sf \rightarrow 12(n-4)=30+(n-4)(n-5)$

$\sf \rightarrow 30 = 12(n-4)-(n-4)(n-5)$

$\sf \rightarrow 30 = (n-4)(12-n+5)$

$\rightarrow \sf 30 = (17-n)(n-4)$

$\rightarrow \sf -n^2+21n-68=30$

$\rightarrow \sf n^2+21n+98=0$

$\rightarrow \sf n^2 - 14n-7n+98 = 0$

$\rightarrow \sf (n-14)(n-7)=0$

$\rightarrow \boxed{\bold{n = 14,7}}$

∴ The value of n is 14 or 7.

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