If S_n = 4 + 11 + 21 + 34 + 50 + ... to n terms, then (1 / 60)( S_29 - S_9 ) is equal to

(a) : Given, S_n=4+11+21+34+ * s * s. up to n terms Also, S_n=4+11+21++ up to n terms Subtracting (ii) from (i), we get 0=[4+7+10+13++]-a_n \ Therefore a_n=(n / 2)[2(4)+(n-1)(3)] =(n / 2)[3 n+5]=(3 n^2+5 n / 2) Now, S_n=sum a_n=(3 / 2) sum n^2+(5 / 2) sum n \ =(3 / 2) (n * (n+1)(2 n+1) / 6)+(5 / 2) (n(n+1) / 2) =(n(n+1) / 4)[(2 n+1)+5] \ =(n(n+1) / 4) ((2 n+6) / 2)=(n(n+1)(n+3) / 2) Now, (1 / 60)[S_29-S_9] =(1 / 60)[((29)(30)(32) / 2)-((9)(10)(12) / 2)] =(1 / 60)[13920-540]=(13380 / 60)=223

If S_{n} = 4 + 11 + 21 + 34 + 50 + to n terms, then ... | Sequence & Series (Mathematics)

(a) : Given, $S_n=4+11+21+34+\cdots \cdots$ . up to $n$ terms $\\$ Also, $S_n=4+11+21++$ up to $n$ terms $\\$ Subtracting (ii) from (i), we get $0=[4+7+10+13++]-a_n \\ \therefore a_n=\frac{n}{2}[2(4)+(n-1)(3)] \\=\frac{n}{2}[3 n+5]=\frac{3 n^2+5 n}{2}\\$ Now, $S_n=\sum a_n=\frac{3}{2} \sum n^2+\frac{5}{2} \sum n$ $\\ =\frac{3}{2} \frac{n \cdot(n+1)(2 n+1)}{6}+\frac{5}{2} \frac{n(n+1)}{2} \\=\frac{n(n+1)}{4}[(2 n+1)+5] \\ =\frac{n(n+1)}{4} \frac{(2 n+6)}{2}=\frac{n(n+1)(n+3)}{2}\\$ Now, $\frac{1}{60}\left[S_{29}-S_9\right] \\=\frac{1}{60}\left[\frac{(29)(30)(32)}{2}-\frac{(9)(10)(12)}{2}\right]$ $\\=\frac{1}{60}[13920-540]=\frac{13380}{60}=223$

Explanation

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