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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
$2.{}^{20}{C_0} + 5.{}^{20}{C_1} + 8.{}^{20}{C_2} + 11.{}^{20}{C_3} + ......62.{}^{20}{C_{20}}$ is equal to
  • A
    ${2^{23}}$
  • B
    ${2^{26}}$
  • C
    ${2^{24}}$
  • ✓
    ${2^{25}}$
✓

Answer

Correct option: D.
${2^{25}}$
d
$2 \cdot^{20} \mathrm{C}_{0}+5 \cdot^{20} \mathrm{C}_{1}+8 \cdot^{20} \mathrm{C}_{2}+11 \cdot^{20} \mathrm{C}_{3}+\ldots \ldots+62 \cdot^{20} \mathrm{C}_{20} 2$

$ = \sum\limits_{r = 0}^{20} {(3r + 2){\,^{20}}} {C_r}$

$ = 3\sum\limits_{r = 0}^{20} r { \cdot ^{20}}{C_r} + 2\sum\limits_{r = 0}^{20} {{\,^{20}}} {C_r}$

$ = 3\sum\limits_{r = 0}^{20} {r\left( {\frac{{20}}{r}} \right)} {\,^{19}}{C_{r - 1}} + {2.2^{20}}$

$=60.2^{19}+2.2^{20}=2^{25}$

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