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Joint Entrance Examination
Mathematics
Binomial Theorem
Question 239
Question 239 of 347
Medium
MCQ
The sum of the series
2
⋅
20
C
0
+
5
⋅
20
C
1
+
8
⋅
20
C
2
+
11
⋅
20
C
3
+
⋯
+
62
⋅
20
C
20
2\cdot{}^{20}C_0+5\cdot{}^{20}C_1+8\cdot{}^{20}C_2+11\cdot{}^{20}C_3+\cdots+62\cdot{}^{20}C_{20}
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
26
2^{26}
2
26
B
2
25
2^{25}
2
25
C
2
23
2^{23}
2
23
D
2
24
2^{24}
2
24
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Solutions · 1
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Official solution
\[\sum_{r=0}^{20}(3r+2){}^{20}C_r = 3\sum_{r=0}^{20}r\,{}^{20}C_r+2\sum_{r=0}^{20}{}^{20}C_r = 3\times20\times2^{19}+2\times2^{20}\] \[=60\times2^{19}+2\times2^{20}=2^{19}(60+4)=2^{19}\times64=2^{25}\]
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