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The value of `Sigma_(r=1)^(n) (a+r+ar)(-a)^r` is equal toA. `(-1)^n[n+1)a^(n+1)-a]`B. `(-1)^n(n+1)a^(n+1)`C. `(-1)^n((n+2)a^(n+1))/2`D. `(-1)^n(na^n)/(2)` |
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Answer» Correct Answer - B `T_(r)=r(-a)^(r )+(r+1)a(-a)^(r )` `=r(-a)^(r )-(r+1)(-a)^(r+1)` `=v(r )-v(r+1)` So, `sum_(r=0)^(n)T_(r )=sum_(r=0)^(n)(v(r )-v(r+1))` `=v_(0)-v_(n+1)` `=-(n+1)(-a)^(n+1)` |
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