If -j-01030-j C10-j-m C20-p C21- then

The correct options are A m = 41 B p = 30Given : nbsp;∑_j=0^10 ^30+jC_10+j= ^mC_20 ^pC_21 Now, nbsp; ∑_j=0^10 ^30+jC_10+j = ^30C_10+ ^31C_11+ ^32C_12+⋯+ ...

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Standard XII

Mathematics

Local Maxima

If ∑j=01030+j...

Question

If $10 \sum j = 0^{30 + j} C_{10 + j} =^{m} C_{20} -^{p} C_{21}$ , then

m = 41

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p = 30

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m = 40

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p = 31

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Solution

The correct options are
A $m = 41$
B $p = 30$
Given : $10 \sum j = 0^{30 + j} C_{10 + j} =^{m} C_{20} -^{p} C_{21}$
Now,
$10 \sum j = 0^{30 + j} C_{10 + j} =^{30} C_{10} +^{31} C_{11} +^{32} C_{12} + \dots +^{40} C_{20} =^{30} C_{9} +^{30} C_{10} +^{31} C_{11} + \dots +^{40} C_{20} -^{30} C_{9}$
Using $^{n} C_{r} +^{n} C_{r - 1} =^{n + 1} C_{r}$
$=^{41} C_{20} -^{30} C_{9}$
Therefore,
$^{m} C_{20} -^{p} C_{21} =^{41} C_{20} -^{30} C_{9} \Rightarrow^{m} C_{20} -^{p} C_{21} =^{41} C_{20} -^{30} C_{21} \Rightarrow m = 41, p = 30$

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Local Maxima

Standard XII Mathematics

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If 10∑j=0 30+jC10+j= mC20− pC21, then

The correct options are A m=41 B p=30Given : 10∑j=0 30+jC10+j= mC20− pC21 Now, 10∑j=0 30+jC10+j= 30C10+ 31C11+ 32C12+⋯+ 40C20= 30C9+ 30C10+ 31C11+⋯+ 40C20− 30C9 Using nCr+ nCr−1= n+1Cr = 41C20− 30C9 Therefore, mC20− pC21= 41C20− 30C9⇒ mC20− pC21= 41C20− 30C21⇒m=41, p=30

If $10 \sum j = 0^{30 + j} C_{10 + j} =^{m} C_{20} -^{p} C_{21}$ , then