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If `(1+px+x^(2))^(n)=1+a_(1)x+a_(2)x^(2)+…+a_(2n)x^(2n)`. The remainder obtained when `a_(1)+5a_(2)+9a_(3)+13a_(4)+…+(8n-3)a_(2n)` is divided by `(p+2)` isA. `1`B. `2`C. `3`D. `0` |
Answer» Correct Answer - C `(c )` `a_(1)+5a_(2)+9a_(3)+…+(8n-3)a_(2n)=sum_(r=1)^(2n)(4r-3)a_(r )` `=4sum_(r=1)^(2n)ra_(r )-3sum_(r=1)^(2n)a_(r )` `(1+px+x^(2))^(n)=1+a_(1)x+a_(2)x^(2)+….+a_(2n)X^(2n)` so, `sum_(r=1)^(2n)a_(r )=(p+2)^(n)-1` Differentiating the expansion and substituting `x=1` `sum_(r=1)^(2n)rar_(r)=n(p+2)^(n)` `:.sum_(r=1)^(2n)(4r-3)a_(r )=4n(p+2)^(n)-3((p+2)^(n)-1)` `=(4n-3)(p+2)^(n)+3` |
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