For integer $n>0$, find the values of $$\\sum\\limits_{i=1}^{n}(-1)^{i+1}\\cdot i\\cdot\\binom{n-1}{i-1}$$<\/p>\n\n\n\n
Click here<\/a> for the solution.<\/p>\n Solution: <\/strong>If $n=1$, then<\/p>\n\n\n\n\\begin{flalign*}\n\\sum\\limits_{i=1}^{n}(-1)^{i+1}\\cdot i\\cdot\\binom{n-1}{i-1} &= (-1)^{1+1}\\cdot 1\\cdot\\binom{1-1}{1-1}& \\\\\n&=1&\n\\end{flalign*}\n\n\n\n If $n=2$, then<\/p>\n\n\n\n\\begin{flalign*}\n\\sum\\limits_{i=1}^{n}(-1)^{i+1}\\cdot i\\cdot\\binom{n-1}{i-1} &=(-1)^{1+1}\\cdot 1\\cdot\\binom{2-1}{1-1}+(-1)^{2+1}\\cdot 2\\cdot\\binom{2-1}{2-1}& \\\\\n&=1-2& \\\\\n&=-1&\n\\end{flalign*}\n\n\n\n If $n\\ge 3$, then<\/p>\n\n\n\n\\begin{flalign*}\n\\sum\\limits_{i=1}^{n}(-1)^{i+1}\\cdot i\\cdot\\binom{n-1}{i-1} &= \\sum\\limits_{i=0}^{n-1}(-1)^{i+2}\\cdot (i+1)\\cdot\\binom{n-1}{i} &\\\\\n&=\\sum\\limits_{i=0}^{n-1}(-1)^i\\cdot i\\cdot \\binom{n-1}{i}+\\sum\\limits_{i=0}^{n-1}(-1)^i\\cdot\\binom{n-1}{i}& \\\\\n&=\\sum\\limits_{i=1}^{n-1}(-1)^i\\cdot i\\cdot \\binom{n-1}{i}+(1-1)^{n-1} &\\\\\n&=\\sum\\limits_{i=1}^{n-1}(-1)^i\\cdot i\\cdot \\dfrac{(n-1)!}{i!\\cdot ((n-1)-i)!} + 0 &\\\\\n&=(n-1)\\cdot\\sum\\limits_{i=1}^{n-1}(-1)^i\\cdot \\dfrac{(n-2)!}{(i-1)!\\cdot ((n-2)-(i-1))!} &\\\\\n&=(n-1)\\cdot(-1)\\cdot\\sum\\limits_{i=1}^{n-1}(-1)^{i-1}\\cdot \\binom{n-2}{i-1} &\\\\\n&=(1-n)\\cdot\\sum\\limits_{i=0}^{n-2}(-1)^i\\cdot \\binom{n-2}{i} &\\\\\n&=(1-n)\\cdot(1-1)^{n-2} &\\\\\n&=0 &\n\\end{flalign*}\n\n\n\n Therefore, we have<\/p>\n\n\n\n\\begin{flalign*}\n\\sum\\limits_{i=1}^{n}(-1)^{i+1}\\cdot i\\cdot\\binom{n-1}{i-1}=&\n\\left\\{\n\\begin{array}{rl@{\\qquad}l}\n 1 & n = 1 \\\\\n-1 & n = 2 \\\\\n 0 & n \\ge 3 \\\\\n\\end{array}\n\\right. & \\\\\n\\end{flalign*}\n\n\n\n<\/div>\n","protected":false},"excerpt":{"rendered":" For integer $n>0$, find the values of $$\\sum\\limits_{i=1}^{n}(-1)^{i+1}\\cdot i\\cdot\\binom{n-1}{i-1}$$ Click here for the solution. Solution: If $n=1$, then \\begin{flalign*} \\sum\\limits_{i=1}^{n}(-1)^{i+1}\\cdot i\\cdot\\binom{n-1}{i-1} &= (-1)^{1+1}\\cdot 1\\cdot\\binom{1-1}{1-1}& \\\\ &=1& \\end{flalign*} If $n=2$, then \\begin{flalign*} \\sum\\limits_{i=1}^{n}(-1)^{i+1}\\cdot i\\cdot\\binom{n-1}{i-1} &=(-1)^{1+1}\\cdot 1\\cdot\\binom{2-1}{1-1}+(-1)^{2+1}\\cdot 2\\cdot\\binom{2-1}{2-1}& \\\\ &=1-2& \\\\ &=-1& … Continue reading →<\/span><\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"inline_featured_image":false},"categories":[13],"tags":[],"_links":{"self":[{"href":"https:\/\/mathfun4kids.com\/mlog\/index.php?rest_route=\/wp\/v2\/posts\/4716"}],"collection":[{"href":"https:\/\/mathfun4kids.com\/mlog\/index.php?rest_route=\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/mathfun4kids.com\/mlog\/index.php?rest_route=\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/mathfun4kids.com\/mlog\/index.php?rest_route=\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/mathfun4kids.com\/mlog\/index.php?rest_route=%2Fwp%2Fv2%2Fcomments&post=4716"}],"version-history":[{"count":64,"href":"https:\/\/mathfun4kids.com\/mlog\/index.php?rest_route=\/wp\/v2\/posts\/4716\/revisions"}],"predecessor-version":[{"id":4803,"href":"https:\/\/mathfun4kids.com\/mlog\/index.php?rest_route=\/wp\/v2\/posts\/4716\/revisions\/4803"}],"wp:attachment":[{"href":"https:\/\/mathfun4kids.com\/mlog\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=4716"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/mathfun4kids.com\/mlog\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=4716"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/mathfun4kids.com\/mlog\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=4716"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}