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Strike441 [17]
3 years ago
10

For the given statement Pn, write the statements P1, Pk, and Pk+1.

Mathematics
2 answers:
fredd [130]3 years ago
8 0
P(n):2+4+6+\cdots+2n=\displaystyle\sum_{i=1}^n2i

P(1):\displaystyle\sum_{i=1}^12i=2=1(1+1)=2

P(k):\displaystyle\sum_{i=1}^k2i=2+\cdots+2k=k(k+1)

P(k+1):\displaystyle\sum_{i=1}^{k+1}2i=2+\cdots+2k+2(k+1)=(k+1)(k+2)
Leno4ka [110]3 years ago
7 0

Answer:

P_{1} =  2

P_{k} = k(k+1)

P_{k+1} = (k+1)(k+2)

Step-by-step explanation:

We are given the statement,

P_{n} as 2 + 4 + 6 + . . . + 2n = n(n+1)

That is,

P_{n} as 2 + 4 + 6 + . . . + 2n = \sum_{i=1}^{n}2i

So, we have,

P_{1} = \sum_{i=1}^{1}2i = 2

P_{k} = \sum_{i=1}^{k}2i = 2 + 4 + 6 + . . . + 2k = k(k+1)

P_{k+1} = \sum_{i=1}^{k}2i = 2 + 4 + 6 + . . . + 2k + 2(k+1) = (k+1)(k+2)

Thus, we get,

P_{1} =  2

P_{k} = k(k+1)

P_{k+1} = (k+1)(k+2)

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