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STD 11 - 7. binomial theoram — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEESTD 11 - 7. binomial theoram4 Marks
MCQ
The coefficient of ${x^n}$ in $\frac{{{{(1 + x)}^2}}}{{{{(1 - x)}^3}}}$ is
  • A
    $3{n^2} + 2n + 1$
  • ✓
    $2{n^2} + 2n + 1$
  • C
    ${n^2} + n + 1$
  • D
    $2{n^2} - 2n + 1$
✓

Answer

Correct option: B.
$2{n^2} + 2n + 1$
b
(b) $\frac{{{{(1 + x)}^2}}}{{{{(1 - x)}^3}}} = {(1 + x)^2}{(1 - x)^{ - 3}}$

$ = (1 + 2x + {x^2})[1 + {\,^3}{C_1}x + {\,^4}{C_2}{x^2} + ....$$ + {\,^{.3 + n - 3}}{C_{n - 2}}{x^{n - 2}} + {\,^{3 + n - 2}}{C_{n - 1}}{x^{n - 1}} + {\,^{3 + n - 1}}{C_n}{x^n} + ....]$

 Coefficient of
${x^n}$$ = {\,^{3 + n - 1}}{C_n} + {2.^{3 + n - 2}}{C_{n - 1}}{ + ^{3 + n - 3}}{C_{n - 2}}$

$ = {\,^{2 + n}}{C_n} + {2.^{n + 1}}{C_{n - 1}} + {\,^n}{C_{n - 2}}$

$ = \frac{{(2 + n)!}}{{n!2!}} + \frac{{2.(n + 1)!}}{{(n - 1)!2!}} + \frac{{n!}}{{(n - 2)!2!}}$

$ = \frac{{n![(2 + n)(1 + n) + 2(n + 1)n + n(n - 1)]}}{{n!\,2!}}$

$ = 2{n^2} + 2n + 1$.

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