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Question

Sum to n: 11+12+14+21+22+24+31+32+34

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Solution

let Sn=11+12+14+21+22+24+31+32+34+---n
tn=r1+r2+r4
tn=rr4+r2+1
tn=r(r2+r+1)(r2r+1)
tn=12×2r(r2+r+1)(r2r+1)
=12×[(r2+r+1)(r2r+1)](r2+r+1)(r2r+1)
tn=12[1r2r+11r2+r+1]
Sn=nr=1tn
=12[1121+1112+1+1+1222+1122+2+1+...+1n2n+11n2+n+1
Sn=12[11n2+n+1]
Sn=12(n2+n+11)n2+n+1
Sn=n(n+1)2(n2+n+1)


1076557_1184166_ans_63d8035f518b46bd982ad165c5b1a1e5.png

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