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`(n+2)^n C_0 2^(n+1)-(n+1)^n C_1 2^n^n C_2 2^(n-1)-`is equal to`4`b. `4n`c. `4(n+1)`d. `2(n+2)`A. 4B. 4nC. 4(n+1)D. 2(n+2) |
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Answer» Correct Answer - C `t_(r+1)=(-1)^(r)(n-r+2).^(n)C_(r)2^(n-r+1)` `= (n+2)2^(n+1)(-1)^(r).^(n)C_(r)(1/2)^(r)-2^(n+1)r.^(n)C_(r)(1/2)^(r)` `= (n+2)2^(n+1).^(n)C_(r)(-1/2)^(r) + 2^(n)n.^(n-1)C_(r-1)(-1/2)^(r-1)` `:.` Sum `= (n+2)2^(n+1){.^(n)C_(0) - .^(n)C_(1) xx 1/2 .^(n)C_(2)xx(1/2)^(2)-"...."}` `+2^(n){.^(n-1)C_(0)-.^(n-1)C_(1)xx1/2+.^(n-1)C_(2)xx(1/2)^(2)+"...."}` `= (n+2)2^(n+1)(1-1/2)^(n)+n2^(n)(1-1/2)^(n-1)` `= 2(n+2)+2n` `= 4n+4` |
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