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!"#