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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
