Question #b2691

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Answer 1 · 2016-09-20T19:14:59

$"Sum="736$ .

Let $S$ be the sum. Then, rearranging the terms of the series,

$S=[1+4+7+10+13+... ]+[3+6+9+12+... +45]$

#=U+V, say, where,

$U=[1+4+7+10+13+... ], &,$

$V=[3+6+9+12+... +45]$

We notice that the General $n^(th)$ Term $u_n$ of the Series $U$ is,

$u_n=1+(n-1)3=3n-2", whereas, that for V is "v_n=3n$ .

Clearly, $u_16=46, and, v_15=45$ .

Therefore, $U=sum_(n=1)^16u_n=sum_(n=1)^16(3n-2)$

$=3sum_(n=1)^16n-2sum_(n=1)^16 1$

$=3[1/2n(n+1)]_(n=1)^16-2(16)=3/2*16*17-32=408-32=376$

Similarly, $V=360$ .

Finally, therefore, $S=376+360=736.$

In fact, $U$ can easily obtained by using the formula

$U=sum_(n=1)^16u_n=[n/2*"(first term+last term)"]$

$=16/2(1+46)=8*47=376, "as before!"$