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Q.Sum the series 5+11+19+29+41+…5 + 11 + 19 + 29 + 41 + \ldots to n terms.

Bihar BsebBihar Board Intermediate 1st Year 2022Subjective· 3mImportance★★★★★
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With tn=n2+3n+1t_n=n^2+3n+1, the sum is n(n+2)(n+4)3\dfrac{n(n+2)(n+4)}{3}.

The series 5,11,19,29,41,…5,11,19,29,41,\dots has first differences 6,8,10,12,…6,8,10,12,\dots and constant second difference 22, so tn=an2+bn+ct_n=an^2+bn+c.

Using t1=5,t2=11,t3=19t_1=5,t_2=11,t_3=19: a+b+c=5a+b+c=5, 4a+2b+c=114a+2b+c=11, 9a+3b+c=199a+3b+c=19. Solving, a=1,b=3,c=1a=1,b=3,c=1, so tn=n2+3n+1t_n=n^2+3n+1.

Now sum:

Sn=∑(n2+3n+1)=n(n+1)(2n+1)6+3⋅n(n+1)2+nS_n=\sum(n^2+3n+1)=\dfrac{n(n+1)(2n+1)}{6}+3\cdot\dfrac{n(n+1)}{2}+n.

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