7.Binomial Theorem
hard

श्रेणी $2 .{ }^{20} C _{0}+5 .{ }^{20} C _{1}+8 .{ }^{20} C _{2}+11 .{ }^{20} C _{3}+\ldots  +62 .{ }^{20} C _{20}$ का योग बराबर है

A

${2^{23}}$

B

${2^{26}}$

C

${2^{24}}$

D

${2^{25}}$

(JEE MAIN-2019)

Solution

$2 \cdot^{20} \mathrm{C}_{0}+5 \cdot^{20} \mathrm{C}_{1}+8 \cdot^{20} \mathrm{C}_{2}+11 \cdot^{20} \mathrm{C}_{3}+\ldots \ldots+62 \cdot^{20} \mathrm{C}_{20} 2$

$ = \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}}$

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

Standard 11
Mathematics

Similar Questions

Start a Free Trial Now

Confusing about what to choose? Our team will schedule a demo shortly.