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Question Number 65199 by rajesh4661kumar@gmail.com last updated on 26/Jul/19

	Answered by Tanmay chaudhury last updated on 26/Jul/19
	$(1),(2,3,4),(5,6,7,8,9),(10,11,12,13,14,15,16)... 1st group contains→ 1 terms 2nd group contains → 3 terms 3rd group contains→ 5terms 4th grouo contains → 7terms ..... ..... nth group contains →{1+(n−1)2}→ 2n−1 terms determination of first term of nth group let t_n =be the first term of nth group taking the first term of each group S=1+2+5+10+17+....+t_n (←total nterms) S= 1+2 +5+10+....+t_(n−1) +t_n (←total n terms) substructing 0=1+1+3+5+7+....−t_n (←total n+1 terms) t_n =1+1+3+5+7.....(←total nterms) t_n =1+(1+3+5+7+...←total n−1 terms) t_n =1+((n−1)/2)[2×1+(n−1−1)×2] t_n =1+((n−1)/2)×(2n−2) t_n =n^2 −2n+2 so sum of terms in nth group is S_(nth) =((2n−1)/2)[2t_n +(2n−1−1)×1] =((2n−1)/2)[2n^2 −4n+4+2n−2] =(2n−1)(n^2 −n+1) =2n^3 −2n^2 +2n−n^2 +n−1 =n^3 +n^3 −3n^2 +3n−1 =(n)^3 +(n−1)^3 proved$
	$(1), (2, 3, 4), (5, 6, 7, 8, 9), (10, 11, 12, 13, 14, 15, 16) . . .$ $1 s t g r o u p c o n t a i n s \to 1 t e r m s$ $2 n d g r o u p c o n t a i n s \to 3 t e r m s$ $3 r d g r o u p c o n t a i n s \to 5 t e r m s$ $4 t h g r o u o c o n t a i n s \to 7 t e r m s$ $. . . . .$ $. . . . .$ $n t h g r o u p c o n t a i n s \to {1 + (n - 1) 2} \to 2 n - 1 t e r m s$ $d e t e r m i n a t i o n o f f i r s t t e r m o f n t h g r o u p$ $l e t t_{n} = b e t h e f i r s t t e r m o f n t h g r o u p$ $t a k i n g t h e f i r s t t e r m o f e a c h g r o u p$ $S = 1 + 2 + 5 + 10 + 17 + . . . . + t_{n} (\leftarrow t o t a l n t e r m s)$ $S = 1 + 2 + 5 + 10 + . . . . + t_{n - 1} + t_{n} (\leftarrow t o t a l n t e r m s)$ $s u b s t r u c t i n g$ $0 = 1 + 1 + 3 + 5 + 7 + . . . . - t_{n} (\leftarrow t o t a l n + 1 t e r m s)$ $t_{n} = 1 + 1 + 3 + 5 + 7 . . . . . (\leftarrow t o t a l n t e r m s)$ $t_{n} = 1 + (1 + 3 + 5 + 7 + . . . \leftarrow t o t a l n - 1 t e r m s)$ $t_{n} = 1 + \frac{n - 1}{2} [2 \times 1 + (n - 1 - 1) \times 2]$ $t_{n} = 1 + \frac{n - 1}{2} \times (2 n - 2)$ $t_{n} = n^{2} - 2 n + 2$ $s o s u m o f t e r m s i n n t h g r o u p i s$ $S_{n t h} = \frac{2 n - 1}{2} [2 t_{n} + (2 n - 1 - 1) \times 1]$ $= \frac{2 n - 1}{2} [2 n^{2} - 4 n + 4 + 2 n - 2]$ $= (2 n - 1) (n^{2} - n + 1)$ $= 2 n^{3} - 2 n^{2} + 2 n - n^{2} + n - 1$ $= n^{3} + n^{3} - 3 n^{2} + 3 n - 1$ $= {(n)}^{3} + {(n - 1)}^{3} p r o v e d$
	Commented by peter frank last updated on 26/Jul/19

	$t h a n k y o u$
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