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Find the value of `^20 C_0-(^(20)C_1)/2+(^(20)C_2)/3-(^(20)C_3)/4+dot` |
Answer» Correct Answer - `1/21` `.^(20)C_(0) - (.^(20)C_(1))/(2) + (.^(20)C_(2))/(3) - (.^(20)C_(3))/(4) + "....."` `= underset(r=0)overset(20)sum(.^(20)C_(r))/(r+1)(-1)^(r)` `= underset(r=0)overset(20)sum(.^(21)C_(r+1))/(20+1)(-1)^(r)` ` = -1/21 underset(r=0)overset(20)sum.^(21)C_(r+1)(-1)^(r+1)` `= - 1/21[-.^(21)C_(1) + .^(21)C_(2)-.^(21)C_(3) + "...."]` `= - 1/(21)[(.^(21)C_(0) - .^(21)C_(1) + .^(21)C_(2) - .^(21)C_(3) + ".....")-.^(21)C_(0) ]` `= -(1)/(21) [(1-1)^(21) - 1]` `= 1/21` |
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