The sum of the series 2 cdot -20 C - 0-5 cdot-20 C - 1-8 cdot -20 C - 2-11 cdot -20 C - 3-cdots -62 -20 C -20is equal to-A- 224B- 225C- 223D- 226

The correct option is B 2^252· ^20C_0+5· ^20C_1+ 8· ^20C_2+11· ^20C_3+ ⋯+62· ^20C_20 =(3·0+2)·^20C_0+(3·1+2)·^20C_1+(3·2+2)·^20C_2 +1(3·3+2)·^20C_3+ ⋯+(3·20+ ...

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Byju's Answer

Standard XII

Mathematics

Sum of Product of Binomial Coefficients

The sum of th...

Question

The sum of the series
$2 \cdot^{20} C_{0} + 5 \cdot^{20} C_{1} + 8 \cdot^{20} C_{2} + 11 \cdot^{20} C_{3} + \dots + 62 \cdot^{20} C_{20}$
is equal to:

2^{24}

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2^{25}

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2^{23}

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2^{26}

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Solution

The correct option is B $2^{25}$
$2 \cdot^{20} C_{0} + 5 \cdot^{20} C_{1} + 8 \cdot^{20} C_{2} + 11 \cdot^{20} C_{3} + \dots + 62 \cdot^{20} C_{20}$
$= (3 \cdot 0 + 2) \cdot^{20} C_{0} + (3 \cdot 1 + 2) \cdot^{20} C_{1} + (3 \cdot 2 + 2) \cdot^{20} C_{2} + 1 (3 \cdot 3 + 2) \cdot^{20} C_{3} + \dots + (3 \cdot 20 + 2) \cdot^{20} C_{20} = 20 \sum r = 0 (3 r + 2)^{20} C_{r} = 3 20 \sum r = 0 r \cdot^{20} C_{r} + 2 \cdot 2^{20}$
$= 3 20 \sum r = 0 r (\frac{20}{r})^{19} C_{r - 1} + 2^{21} = 60 \cdot 2^{19} + 2^{21} = 2^{25}$

Suggest Corrections

The sum of the series 2⋅ 20C0+5⋅ 20C1+8⋅ 20C2+11⋅ 20C3+⋯+62⋅ 20C20 is equal to:

The correct option is B 2252⋅ 20C0+5⋅ 20C1+8⋅ 20C2+11⋅ 20C3+⋯+62⋅ 20C20 =(3⋅0+2)⋅20C0+(3⋅1+2)⋅20C1+(3⋅2+2)⋅20C2 +1(3⋅3+2)⋅20C3+⋯+(3⋅20+2)⋅20C20=20∑r=0(3r+2)20Cr=320∑r=0r⋅20Cr+2⋅220 =320∑r=0r(20r) 19Cr−1+221=60⋅219+221=225

The sum of the series
$2 \cdot^{20} C_{0} + 5 \cdot^{20} C_{1} + 8 \cdot^{20} C_{2} + 11 \cdot^{20} C_{3} + \dots + 62 \cdot^{20} C_{20}$
is equal to: