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Question Number 162074 by naka3546 last updated on 26/Dec/21

Find the exact value of Σ_(k=0) ^(1004) ( _k ^(2014) ) ∙ 3^k .

$${Find}\:\:{the}\:\:{exact}\:\:{value}\:\:{of} \\ $$$$\:\:\:\:\:\:\:\underset{{k}=\mathrm{0}} {\overset{\mathrm{1004}} {\sum}}\:\left(\underset{{k}} {\overset{\mathrm{2014}} {\:}}\right)\:\centerdot\:\mathrm{3}^{{k}} \:. \\ $$

Answered by Rasheed.Sindhi last updated on 26/Dec/21

(a+b)^n =Σ_(k=0) ^(n) ((n),(k) )a^(n−k) b^k Let a=1 , b=3 & n=2014 S(2015)=(1+3)^(2014) =Σ_(k=0) ^(2014) (((2014)),(( k)) ).3^k =4^(2014) ....(i) We′ve to determine first 1005 terms(sum) i-e Σ_(k=0) ^(1004) ( _k ^(2014) ) ∙ 3^k S(1to2015)={t_1 +t_2 +...t_(1007) +(t_(1008) /2)}_(S_1 ) +{(t_(1008) /2)+t_(1009) +t_(1010) +...+t_(2015) }_(S_2 ) ∵Here S_1 =S_2 ∴ S_1 =S/2=4^(2014) /2 ⇒S_1 =4^(2013) ∴ S(1to1005)=S_1 −(t_(1006) +t_(1007) +(t_(1008) /2)) =4^(2013) −( (((2014)),((1005)) ).3^(1005) + (((2014)),((1006)) ).3^(1006) +(1/2) (((2014)),((1007)) ).3^(1007) ) =4^(2013) −3^(1005) ( (((2014)),((1005)) )+3 (((2014)),((1006)) )+(9/2) (((2014)),((1007)) ))

$$\left({a}+{b}\right)^{{n}} =\overset{{n}} {\underset{{k}=\mathrm{0}} {\sum}}\begin{pmatrix}{{n}}\\{{k}}\end{pmatrix}{a}^{{n}−{k}} {b}^{{k}} \\ $$$${Let}\:{a}=\mathrm{1}\:,\:{b}=\mathrm{3}\:\&\:{n}=\mathrm{2014} \\ $$$${S}\left(\mathrm{2015}\right)=\left(\mathrm{1}+\mathrm{3}\right)^{\mathrm{2014}} =\underset{{k}=\mathrm{0}} {\overset{\mathrm{2014}} {\sum}}\begin{pmatrix}{\mathrm{2014}}\\{\:\:\:\:{k}}\end{pmatrix}.\mathrm{3}^{{k}} =\mathrm{4}^{\mathrm{2014}} ....\left({i}\right) \\ $$$${We}'{ve}\:{to}\:{determine}\:{first}\:\mathrm{1005}\:{terms}\left({sum}\right) \\ $$$${i}-{e}\:\underset{{k}=\mathrm{0}} {\overset{\mathrm{1004}} {\sum}}\:\left(\underset{{k}} {\overset{\mathrm{2014}} {\:}}\right)\:\centerdot\:\mathrm{3}^{{k}} \: \\ $$$${S}\left(\mathrm{1to2015}\right)=\underset{\mathrm{S}_{\mathrm{1}} } {\underbrace{\left\{\mathrm{t}_{\mathrm{1}} +\mathrm{t}_{\mathrm{2}} +...\mathrm{t}_{\mathrm{1007}} +\frac{\mathrm{t}_{\mathrm{1008}} }{\mathrm{2}}\right\}}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\underset{{S}_{\mathrm{2}} } {\underbrace{\left\{\frac{\mathrm{t}_{\mathrm{1008}} }{\mathrm{2}}+\mathrm{t}_{\mathrm{1009}} +\mathrm{t}_{\mathrm{1010}} +...+\mathrm{t}_{\mathrm{2015}} \right\}}} \\ $$$$\:\because{Here}\:{S}_{\mathrm{1}} ={S}_{\mathrm{2}} \:\:\:\:\therefore\:{S}_{\mathrm{1}} ={S}/\mathrm{2}=\mathrm{4}^{\mathrm{2014}} /\mathrm{2} \\ $$$$\Rightarrow{S}_{\mathrm{1}} =\mathrm{4}^{\mathrm{2013}} \\ $$$$\therefore\:{S}\left(\mathrm{1to1005}\right)={S}_{\mathrm{1}} −\left(\mathrm{t}_{\mathrm{1006}} +\mathrm{t}_{\mathrm{1007}} +\frac{\mathrm{t}_{\mathrm{1008}} }{\mathrm{2}}\right) \\ $$$$=\mathrm{4}^{\mathrm{2013}} −\left(\begin{pmatrix}{\mathrm{2014}}\\{\mathrm{1005}}\end{pmatrix}.\mathrm{3}^{\mathrm{1005}} +\begin{pmatrix}{\mathrm{2014}}\\{\mathrm{1006}}\end{pmatrix}.\mathrm{3}^{\mathrm{1006}} +\frac{\mathrm{1}}{\mathrm{2}}\begin{pmatrix}{\mathrm{2014}}\\{\mathrm{1007}}\end{pmatrix}.\mathrm{3}^{\mathrm{1007}} \right) \\ $$$$=\mathrm{4}^{\mathrm{2013}} −\mathrm{3}^{\mathrm{1005}} \left(\begin{pmatrix}{\mathrm{2014}}\\{\mathrm{1005}}\end{pmatrix}+\mathrm{3}\begin{pmatrix}{\mathrm{2014}}\\{\mathrm{1006}}\end{pmatrix}+\frac{\mathrm{9}}{\mathrm{2}}\begin{pmatrix}{\mathrm{2014}}\\{\mathrm{1007}}\end{pmatrix}\right) \\ $$

Commented by naka3546 last updated on 26/Dec/21

Thank you, sir.

$${Thank}\:\:{you},\:\:{sir}. \\ $$

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