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Q.Find the sum of nn terms of the series 1×2×3+2×3×4+3×4×5+⋯1\times2\times3+2\times3\times4+3\times4\times5+\cdots OR Find the sum of nn terms of the sequence 8,88,888,8888,…8,88,888,8888,\ldots

Rajasthan RbseRajasthan Board Senior Secondary Part-I Examination 2023Subjective· 6mImportance★★★★★
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1×2×3+2×3×4+⋯1\times2\times3+2\times3\times4+\cdots up to nn terms sums to n(n+1)(n+2)(n+3)4\dfrac{n(n+1)(n+2)(n+3)}{4}.

We solve the primary question (the OR alternative, on the sequence 8,88,888,…8,88,888,\ldots, is not needed since this one is fully answerable).

The rr-th term of the series is Tr=r(r+1)(r+2)T_r=r(r+1)(r+2). Expanding:

Tr=r(r2+3r+2)=r3+3r2+2rT_r = r(r^2+3r+2) = r^3+3r^2+2r

Summing from r=1r=1 to nn, using the standard formulas ∑r3=(n(n+1)2)2\sum r^3=\left(\dfrac{n(n+1)}{2}\right)^2, ∑r2=n(n+1)(2n+1)6\sum r^2=\dfrac{n(n+1)(2n+1)}{6}, ∑r=n(n+1)2\sum r=\dfrac{n(n+1)}{2}:

Sn=∑r=1nTr=(n(n+1)2)2+3⋅n(n+1)(2n+1)6+2⋅n(n+1)2S_n = \sum_{r=1}^n T_r = \left(\frac{n(n+1)}{2}\right)^2 + 3\cdot\frac{n(n+1)(2n+1)}{6} + 2\cdot\frac{n(n+1)}{2}

=n2(n+1)24+n(n+1)(2n+1)2+n(n+1)= \frac{n^2(n+1)^2}{4} + \frac{n(n+1)(2n+1)}{2} + n(n+1)

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